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Bài 09:
\[P = \left( {\frac{{4\sqrt x }}{{2 + \sqrt x }} + \frac{{8x}}{{4 - x}}} \right):\left( {\frac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \frac{2}{{\sqrt x }}} \right)\]
*Điều kiện xác định:
\[\left\{ \begin{array}{l}
\sqrt x \ge 0\\
2 + \sqrt x \ne 0\\
4 - x \ne 0\\
x - 2\sqrt x \ne 0\\
\sqrt x \ne 0\\
\frac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \frac{2}{{\sqrt x }} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
\sqrt x \left( {\sqrt x - 2} \right) \ne 0\\
x \ne 0\\
\frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \frac{2}{{\sqrt x }} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
\frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \frac{{2\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
\frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \frac{{2\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
\frac{{\sqrt x - 1 - 2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
\frac{{3 - \sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
3 - \sqrt x \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
\sqrt x \ne 3
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
x \ne {3^2}
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\]
*Rút gọn \[P\]
\[P = \left( {\frac{{4\sqrt x }}{{2 + \sqrt x }} + \frac{{8x}}{{4 - x}}} \right):\left( {\frac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \frac{2}{{\sqrt x }}} \right)\]
\[ \Leftrightarrow P = \left[ {\frac{{4\sqrt x \left( {2 - \sqrt x } \right)}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}} + \frac{{8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}} \right]:\left[ {\frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \frac{{2\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]\]
\[ \Leftrightarrow P = \left[ {\frac{{8\sqrt x - 4x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}} + \frac{{8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}} \right]:\left[ {\frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \frac{{2\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]\]
\[ \Leftrightarrow P = \frac{{8\sqrt x - 4x + 8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}:\frac{{\sqrt x - 1 - 2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\]
\[ \Leftrightarrow P = \frac{{8\sqrt x + 4x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}:\frac{{\sqrt x - 3}}{{\sqrt x \left( {2 - \sqrt x } \right)}}\]
\[ \Leftrightarrow P = \frac{{8\sqrt x + 4x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}.\frac{{\sqrt x \left( {2 - \sqrt x } \right)}}{{\sqrt x - 3}}\]
\[ \Leftrightarrow P = \frac{{\sqrt x \left( {8\sqrt x + 4x} \right)}}{{\left( {2 + \sqrt x } \right)\left( {\sqrt x - 3} \right)}}\]
\[ \Leftrightarrow P = \frac{{4x\left( {2 + \sqrt x } \right)}}{{\left( {2 + \sqrt x } \right)\left( {\sqrt x - 3} \right)}}\]
\[ \Leftrightarrow P = \frac{{4x}}{{\sqrt x - 3}}\]
Vậy \[P = \frac{{4x}}{{\sqrt x - 3}}\forall x \in \left\{ {x \in R\left| {x > 0;x \ne 4;x \ne 9} \right.} \right\}\]
*Tìm \[x\] để \[P = 1\]
Ta có:
\[\left\{ \begin{array}{l}
P = \frac{{4x}}{{\sqrt x - 3}}\\
P = 1
\end{array} \right.\]
\[ \Leftrightarrow \frac{{4x}}{{\sqrt x - 3}} = 1\]
\[ \Leftrightarrow 4x = \sqrt x - 3 - 4x\]
\[ \Leftrightarrow - 4x + \sqrt x - 3 = 0\]
\[ \Leftrightarrow - \left[ {{{\left( {2\sqrt x } \right)}^2} - 2.2.\frac{1}{4}\sqrt x + {{\left( {\frac{1}{4}} \right)}^2} + \frac{{47}}{{16}}} \right] = 0\]
\[ \Leftrightarrow - {\left( {2\sqrt x - \frac{1}{4}} \right)^2} - \frac{{47}}{{16}} = 0\]
\[ \Leftrightarrow {\left( {2\sqrt x - \frac{1}{4}} \right)^2} + \frac{{47}}{{16}} = 0\]
\[ \Leftrightarrow x \in \emptyset \]
*Tìm \[m\] để với mọi \[x > 9\] ta có \[m\left( {\sqrt x - 3} \right)P > x + 1\]
Thay \[P = \frac{{4x}}{{\sqrt x - 3}}\] vào \[m\left( {\sqrt x - 3} \right)P > x + 1\]
\[ \Leftrightarrow m\left( {\sqrt x - 3} \right)\frac{{4x}}{{\sqrt x - 3}} > x + 1\]
Ta có:
\[x > 9\]
\[ \Leftrightarrow 4xm > x + 1 > 9 + 1 = 10\]
\[ \Leftrightarrow 4xm > 10\]
\[ \Leftrightarrow 36m > 10\]
\[ \Leftrightarrow m > \frac{{10}}{{36}} = \frac{5}{{18}}\]
\[ \Leftrightarrow m\left( {\sqrt x - 3} \right)P > x + 1\forall \left\{ \begin{array}{l}
m \in \left\{ {m \in R\left| {m > \frac{5}{{18}}} \right.} \right\}\\
x \in \left\{ {x \in R\left| {x > 9} \right.} \right\}\\
P = \left( {\frac{{4\sqrt x }}{{2 + \sqrt x }} + \frac{{8x}}{{4 - x}}} \right):\left( {\frac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \frac{2}{{\sqrt x }}} \right)
\end{array} \right.\]
Vậy 
\[m\left( {\sqrt x - 3} \right)P > x + 1\forall \left\{ \begin{array}{l}
m \in \left\{ {m \in R\left| {m > \frac{5}{{18}}} \right.} \right\}\\
x \in \left\{ {x \in R\left| {x > 9} \right.} \right\}\\
P = \left( {\frac{{4\sqrt x }}{{2 + \sqrt x }} + \frac{{8x}}{{4 - x}}} \right):\left( {\frac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \frac{2}{{\sqrt x }}} \right)
\end{array} \right.\]

Bài 04:
\[A = \frac{{\sqrt {1 - \sqrt {1 - {x^2}} } \left[ {\sqrt {{{\left( {1 + x} \right)}^3}} + \sqrt {{{\left( {1 - x} \right)}^3}} } \right]}}{{2 - \sqrt {1 - {x^2}} }}\]
Điều kiện xác định:
\[\left\{ \begin{array}{l}
1 - \sqrt {1 - {x^2}} \ge 0\\
1 + x \ge 0\\
1 - x \ge 0\\
2 - \sqrt {1 - {x^2}} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {1 - {x^2}} \le 1\\
x \ge - 1\\
x \le 1\\
\sqrt {1 - {x^2}} \ne 2
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
1 - {x^2} \le 1\\
- 1 \le x \le 1\\
1 - {x^2} \ne 4
\end{array} \right.\]
\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{x^2} \ge 0}\\
{ - 1 \le x \le 1}\\
{{x^2} \ne - 3}
\end{array}} \right.\]
\[ \Leftrightarrow - 1 < x \le 1\]

\[ \Leftrightarrow A = \frac{{\sqrt {\left( {1 - \sqrt {1 - {x^2}} } \right){{\left( {1 + x} \right)}^3}} + \sqrt {\left( {1 - \sqrt {1 - {x^2}} } \right){{\left( {1 - x} \right)}^3}} }}{{2 - \sqrt {1 - {x^2}} }}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left( {2 - 2\sqrt {1 - {x^2}} } \right){{\left( {1 + x} \right)}^3}} + \sqrt {\left( {2 - 2\sqrt {1 - {x^2}} } \right){{\left( {1 - x} \right)}^3}} }}{{\sqrt 2 \left( {2 - \sqrt {1 - {x^2}} } \right)}}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left[ {2 - 2\sqrt {\left( {1 - x} \right)\left( {1 + x} \right)} } \right]{{\left( {1 + x} \right)}^3}} + \sqrt {\left[ {2 - 2\sqrt {\left( {1 - x} \right)\left( {1 + x} \right)} } \right]{{\left( {1 - x} \right)}^3}} }}{{\sqrt 2 \left( {2 - \sqrt {1 - {x^2}} } \right)}}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left[ {{{\left( {\sqrt {1 + x} } \right)}^2} - 2\sqrt {\left( {1 - x} \right)\left( {1 + x} \right)} + {{\left( {\sqrt {1 - x} } \right)}^2}} \right]{{\left( {1 + x} \right)}^3}} + \sqrt {\left[ {{{\left( {\sqrt {1 + x} } \right)}^2} - 2\sqrt {\left( {1 - x} \right)\left( {1 + x} \right)} + {{\left( {\sqrt {1 - x} } \right)}^2}} \right]{{\left( {1 - x} \right)}^3}} }}{{\sqrt 2 \left[ {2 - \sqrt {\left( {1 - x} \right)\left( {1 + x} \right)} } \right]}}\]
\[ \Leftrightarrow A = \frac{{\left| {\sqrt {1 + x} - \sqrt {1 - x} } \right|\left| {1 + x} \right|\sqrt {1 + x} + \left| {\sqrt {1 + x} - \sqrt {1 - x} } \right|\left| {1 - x} \right|\sqrt {1 - x} }}{{2\sqrt 2 - \sqrt {2\left( {1 - x} \right)\left( {1 + x} \right)} }}\]
\[ \Leftrightarrow A = \frac{{\left( {\sqrt {1 + x} - \sqrt {1 - x} } \right)\left( {1 + x} \right)\sqrt {1 + x} + \left( {\sqrt {1 + x} - \sqrt {1 - x} } \right)\left( {1 - x} \right)\sqrt {1 - x} }}{{2\sqrt 2 - \sqrt {2\left( {1 - x} \right)\left( {1 + x} \right)} }}\]
\[ \Leftrightarrow A = \frac{{\left( {\sqrt {1 + x} - \sqrt {1 - x} } \right)\left[ {\sqrt {{{\left( {1 + x} \right)}^3}} + \sqrt {{{\left( {1 - x} \right)}^3}} } \right]}}{{2\sqrt 2 - \sqrt {2\left( {1 - x} \right)\left( {1 + x} \right)} }}\]
\[ \Leftrightarrow A = \frac{{\left( {\sqrt {1 + x} - \sqrt {1 - x} } \right)\left( {\sqrt {1 + x} + \sqrt {1 - x} } \right)\left[ {\left( {1 + x} \right) - \sqrt {1 - {x^2}} + \left( {1 - x} \right)} \right]}}{{2\sqrt 2 - \sqrt {2\left( {1 - x} \right)\left( {1 + x} \right)} }}\]
\[ \Leftrightarrow A = \frac{{\left( {1 + x - 1 + x} \right)\left( {2 - \sqrt {1 - {x^2}} } \right)}}{{\sqrt 2 \left( {2 - \sqrt {1 - {x^2}} } \right)}}\]
\[ \Leftrightarrow A = 2x\]

Bài 03:
Rút gọn: \[Q = \frac{{\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } }}{{\sqrt {x + \sqrt {2x - 1} } - \sqrt {x - \sqrt {2x - 1} } }}\]
\[ \Leftrightarrow Q = \frac{{\sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}{{\sqrt {x + \sqrt {2x - 1} } - \sqrt {x - \sqrt {2x - 1} } }}\]
\[ \Leftrightarrow Q = \frac{{\sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}{{\sqrt {\frac{{2x - 1}}{2} + \frac{{2\sqrt {2x - 1} }}{2} + \frac{1}{2}} - \sqrt {\frac{{2x - 1}}{2} - \frac{{2\sqrt {2x - 1} }}{2} + \frac{1}{2}} }}\]
\[ \Leftrightarrow Q = \frac{{\sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}{{\sqrt {\frac{{{{\left( {\sqrt {2x - 1} + 1} \right)}^2}}}{2}} - \sqrt {\frac{{{{\left( {\sqrt {2x - 1} + 1} \right)}^2}}}{2}} }}\]
\[ \Leftrightarrow Q = \frac{{\sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}{{\frac{{\sqrt {{{\left( {\sqrt {2x - 1} + 1} \right)}^2}} }}{{\sqrt 2 }} - \frac{{\sqrt {{{\left( {\sqrt {2x - 1} - 1} \right)}^2}} }}{{\sqrt 2 }}}}\]
*Điều kiện xác định:
\[\left\{ \begin{array}{l}
x - 1 \ge 0\\
2x - 1 \ge 0\\
\frac{{\sqrt {{{\left( {\sqrt {2x - 1} + 1} \right)}^2}} }}{{\sqrt 2 }} - \frac{{\sqrt {{{\left( {\sqrt {2x - 1} - 1} \right)}^2}} }}{{\sqrt 2 }} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \ne \frac{1}{2}
\end{array} \right.\]
\[ \Leftrightarrow x \ge 1\]

\[ \Leftrightarrow Q = \frac{{\left| {\sqrt {x - 1} + 1} \right| + \left| {\sqrt {x - 1} - 1} \right|}}{{\frac{{\left| {\sqrt {2x - 1} + 1} \right| - \left| {\sqrt {2x - 1} - 1} \right|}}{{\sqrt 2 }}}}\]
*Xét trường hợp \[1 \le x < 2\]:
\[ \Leftrightarrow Q = \frac{{\sqrt {x - 1} + 1 - \sqrt {x - 1} + 1}}{{\frac{{\sqrt {2x - 1} + 1 - \sqrt {2x - 1} + 1}}{{\sqrt 2 }}}}\]
\[ \Leftrightarrow Q = 2.\frac{{\sqrt 2 }}{2}\]
\[ \Leftrightarrow Q = \sqrt 2 \]
*Xét trường hợp \[2 \le x\]:
\[ \Leftrightarrow Q = \frac{{\sqrt {x - 1} + 1 + \sqrt {x - 1} - 1}}{{\frac{{\sqrt {2x - 1} + 1 - \sqrt {2x - 1} + 1}}{{\sqrt 2 }}}}\]
\[ \Leftrightarrow Q = \frac{{2\sqrt {x - 1} }}{{\frac{2}{{\sqrt 2 }}}}\]
\[ \Leftrightarrow Q = \sqrt {2x - 2} \]

Bài 10:
\[A = \frac{{\sqrt {x - 1 - 2\sqrt {x - 2} } }}{{\sqrt {x - 2} - 1}}\]

a) Tìm điều kiện cho \[A\]
b) Rút gọn \[A\]
c) Tính \[A\] khi \[x = 6\]




a) Tìm điều kiện cho \[A\]
Ta có:
\[A = \frac{{\sqrt {x - 1 - 2\sqrt {x - 2} } }}{{\sqrt {x - 2} - 1}}\]
\[ \Leftrightarrow A = \frac{{\sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} }}{{\sqrt {x - 2} - 1}}\]
Điều kiện xác định:
\[\left\{ \begin{array}{l}
x - 2 \ge 0\\
\sqrt {x - 2} - 1 \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \ne 3
\end{array} \right.\]




b) Rút gọn \[A\]
\[A = \frac{{\sqrt {x - 1 - 2\sqrt {x - 2} } }}{{\sqrt {x - 2} - 1}}\]
\[ \Leftrightarrow A = \frac{{\sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} }}{{\sqrt {x - 2} - 1}}\]
\[ \Leftrightarrow A = \frac{{\left| {\sqrt {x - 2} - 1} \right|}}{{\sqrt {x - 2} - 1}}\]
*Xét trường hợp \[2 \le x < 3\]:
\[ \Leftrightarrow A = \frac{{1 - \sqrt {x - 2} }}{{\sqrt {x - 2} - 1}}\]
\[ \Leftrightarrow A = - \frac{{\sqrt {x - 2} - 1}}{{\sqrt {x - 2} - 1}} = - 1\]
*Xét trường hợp \[3 < x\]:
\[ \Leftrightarrow A = \frac{{\sqrt {x - 2} - 1}}{{\sqrt {x - 2} - 1}}\]
\[ \Leftrightarrow A = \frac{{\sqrt {x - 2} - 1}}{{\sqrt {x - 2} - 1}} = 1\]




c) Tính \[A\] khi \[x = 6\]
Thay \[x = 6\] vào \[A\], ta có:
\[A = \frac{{\left| {\sqrt {6 - 2} - 1} \right|}}{{\sqrt {6 - 2} - 1}} = 1\]

Bài 08:
\[A = \sqrt {\frac{b}{a}} - \frac{{\sqrt {ab} - \sqrt {{a^2}} }}{a}\]

a)Tìm điều kiện đối với \[a , b\] để \[A\] có nghĩa
\[\left\{ \begin{array}{l}
\frac{b}{a} \ge 0\\
ab \ge 0\\
a \ne 0
\end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\frac{b}{a} \ge 0\\
a \ne 0
\end{array} \right.\\
\left\{ \begin{array}{l}
ab \ge 0\\
a \ne 0
\end{array} \right.
\end{array} \right.\]

b) Rút gọn \[A\]
\[A = \sqrt {\frac{b}{a}} - \frac{{\sqrt {ab} - \sqrt {{a^2}} }}{a}\]
*Xét trường hợp \[a > 0\] và \[b \ge 0\]:
\[ \Leftrightarrow A = \frac{{\sqrt b }}{{\sqrt a }} - \frac{{\sqrt {ab} - a}}{a}\]
\[ \Leftrightarrow A = \frac{{\sqrt {ab} - \sqrt {ab} + a}}{a}\]
\[ \Leftrightarrow A = \frac{a}{a}\]
\[ \Leftrightarrow A = 1\]
*Xét trường hợp \[a < 0\] và \[b \le 0\]:
\[ \Leftrightarrow A = \frac{{\sqrt {\left( { - b} \right)} }}{{\sqrt {\left( { - a} \right)} }} - \frac{{\sqrt {\left( { - a} \right)\left( { - b} \right)} + \sqrt {{{\left( { - a} \right)}^2}} }}{{ - \sqrt {{{\left( { - a} \right)}^2}} }}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left( { - b} \right)} }}{{\sqrt {\left( { - a} \right)} }} - \frac{{\sqrt {\left( { - a} \right)\left( { - b} \right)} + \sqrt {{{\left( { - a} \right)}^2}} }}{{ - \sqrt {{{\left( { - a} \right)}^2}} }}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left( { - b} \right)} }}{{\sqrt {\left( { - a} \right)} }} - \frac{{\sqrt {\left( { - a} \right)} \left[ {\sqrt {\left( { - b} \right)} + \sqrt {\left( { - a} \right)} } \right]}}{{ - \sqrt {{{\left( { - a} \right)}^2}} }}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left( { - b} \right)} }}{{\sqrt {\left( { - a} \right)} }} + \frac{{\sqrt {\left( { - b} \right)} + \sqrt {\left( { - a} \right)} }}{{\sqrt {\left( { - a} \right)} }}\]
\[ \Leftrightarrow A = \frac{{\sqrt {\left( { - a} \right)} - 2\sqrt {\left( { - b} \right)} }}{{\sqrt {\left( { - a} \right)} }}\]

Bài 06:
\[C = \left| {\frac{{x + y}}{2} - \sqrt {xy} } \right| + \left| {\frac{{x + y}}{2} - \sqrt {xy} } \right| - \left| {\frac{x}{3}} \right| - \left| {\frac{y}{3}} \right|\]

Điều kiện xác định:
\[\left\{ \begin{array}{l}
x \ge 0\\
y \ge 0
\end{array} \right.\]

a) Rút gọn \[C\]
\[C = \left| {\frac{{x + y}}{2} - \sqrt {xy} } \right| + \left| {\frac{{x + y}}{2} - \sqrt {xy} } \right| - \left| {\frac{x}{3}} \right| - \left| {\frac{y}{3}} \right|\]
\[ \Leftrightarrow C = \left| {\frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}}{2}} \right| + \left| {\frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{2}} \right| - \left| {\frac{x}{3}} \right| - \left| {\frac{y}{3}} \right|\]
\[ \Leftrightarrow C = \frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}}{2} + \frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{2} - \frac{x}{3} - \frac{y}{3}\]
\[ \Leftrightarrow C = \frac{{{{\left( {\sqrt x - \sqrt y } \right)}^2} + {{\left( {\sqrt x + \sqrt y } \right)}^2}}}{2} - \frac{{x + y}}{3}\]
\[ \Leftrightarrow C = \frac{{x - 2\sqrt {xy} + y + x + 2\sqrt {xy} + y}}{2} - \frac{{x + y}}{3}\]
\[ \Leftrightarrow C = \frac{{2\left( {x + y} \right)}}{2} - \frac{{x + y}}{3}\]
\[ \Leftrightarrow C = \left( {x + y} \right) - \frac{{x + y}}{3}\]
\[ \Leftrightarrow C = \left( {x + y} \right)\left( {1 - \frac{1}{3}} \right)\]
\[ \Leftrightarrow C = \frac{2}{3}\left( {x + y} \right)\]

b) Tìm \[x , y\] biết \[C = \left| {\frac{x}{{2014}}} \right| + \left| {\frac{y}{{2014}}} \right|\]
Ta có:
\[C = \frac{2}{3}\left( {x + y} \right)\]    (chứng minh câu a)
\[ \Leftrightarrow C = \frac{2}{3}\left( {x + y} \right) = \left| {\frac{x}{{2014}}} \right| + \left| {\frac{y}{{2014}}} \right|\]
\[ \Leftrightarrow C = \frac{2}{3}\left( {x + y} \right) = \frac{x}{{2014}} + \frac{y}{{2014}}\]
Mà \[\frac{2}{3} \ne \frac{1}{{2014}}\]
\[ \Leftrightarrow C = \frac{2}{3}\left( {x + y} \right) = \frac{1}{{2014}}\left( {x + y} \right) = 0\]
Vì:
\[\left\{ \begin{array}{l}
\frac{2}{3} \ne 0\\
\frac{1}{{2014}} \ne 0
\end{array} \right.\]
\[ \Leftrightarrow x + y = 0\]
Mà 
\[\left\{ \begin{array}{l}
x \ge 0\\
y \ge 0
\end{array} \right.\]    (điều kiện xác định câu a)
\[ \Leftrightarrow x = y = 0\]

Bài 05:
Tìm giá trị nhỏ nhất của biểu thức: \[B = \sqrt {1 + 4x + 4{x^2}} + \sqrt {4{x^2} - 12x + 9} \]
Ta có:
\[B = \sqrt {1 + 4x + 4{x^2}} + \sqrt {4{x^2} - 12x + 9} \]
\[ \Leftrightarrow B = \sqrt {{{\left( {1 + 2x} \right)}^2}} + \sqrt {{{\left( {2x - 3} \right)}^2}} \]
\[ \Leftrightarrow B = \left| {1 + 2x} \right| + \left| {2x - 3} \right|\]
Ta có:
\[\left\{ \begin{array}{l}
\left| {1 + 2x} \right| \ge 0\forall x \in R\\
\left| {2x - 3} \right| \ge 0 \in \forall xR
\end{array} \right.\]
\[ \Leftrightarrow B = \left| {1 + 2x} \right| + \left| {2x - 3} \right| \ge 0 \in \forall xR\]
\[ \Leftrightarrow Min\left( B \right) = \left| {1 + 2x} \right| + \left| {2x - 3} \right| = 0\] khi \[\left| {1 + 2x} \right| = \left| {2x - 3} \right| = 0\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
\left| {1 + 2x} \right| = 0\\
\left| {2x - 3} \right| = 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
1 + 2x = 0\\
2x - 3 = 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x = - \frac{1}{2}\\
x = \frac{3}{2}
\end{array} \right.\]
Mà \[ - \frac{1}{2} \ne \frac{3}{2}\] nên đẳng thức không xảy ra

Bài 02:
a) Chứng minh \[\sqrt {3 - 4x} + \sqrt {4x + 1} \ge 2\forall x \in \left[ { - \frac{1}{4};\frac{3}{4}} \right]\]
Điều kiện xác định:
\[\left\{ \begin{array}{l}
3 - 4x \ge 0\\
4x + 1 \ge 0
\end{array} \right.\]
\[ \Rightarrow \left\{ \begin{array}{l}
3 \ge 4x\\
4x \ge - 1
\end{array} \right.\]
\[ \Rightarrow \left\{ \begin{array}{l}
x \le \frac{3}{4}\\
x \ge - \frac{1}{4}
\end{array} \right.\]
\[ \Rightarrow - \frac{1}{4} \le x \le \frac{3}{4}\]

\[ \Leftrightarrow 3 - 4x + 4x + 1 + 2\sqrt {3 - 4x} \sqrt {4x + 1} \ge 4\]
\[ \Leftrightarrow 4 + 2\sqrt {12x + 3 - 16{x^2} - 4x} \ge 4\]
\[ \Leftrightarrow 4 + 2\sqrt { - 16{x^2} + 8x + 3} \ge 4\]
\[ \Leftrightarrow \sqrt { - 16{x^2} + 8x + 3} \ge 0\]
\[ \Leftrightarrow - 16{x^2} + 8x + 3 \ge 0\]
\[ \Leftrightarrow - \left( {16{x^2} - 8x - 3} \right) \ge 0\]
\[ \Leftrightarrow - \left[ {{{\left( {4x} \right)}^2} - 2.4x + 1 - 4} \right] \ge 0\]
\[ \Leftrightarrow - \left[ {{{\left( {4x - 1} \right)}^2} - 4} \right] \ge 0\]
\[ \Leftrightarrow - \left[ {{{\left( {4x - 1} \right)}^2} - {{\left( 2 \right)}^2}} \right] \ge 0\]
\[ \Leftrightarrow - \left( {4x - 1 + 2} \right)\left( {4x - 1 - 2} \right) \ge 0\]
\[ \Leftrightarrow - \left( {4x + 1} \right)\left( {4x - 3} \right) \ge 0\]
\[ \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4x + 1 \ge 0\\
4x - 3 \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4x + 1 \le 0\\
4x - 3 \ge 0
\end{array} \right.
\end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4x \ge - 1\\
4x \le 3
\end{array} \right.\\
\left\{ \begin{array}{l}
4x \le - 1\\
4x \ge 3
\end{array} \right.
\end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - \frac{1}{4}\\
x \le \frac{3}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge \frac{3}{4}\\
x \le - \frac{1}{4}
\end{array} \right.
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge - \frac{1}{4}\\
x \le \frac{3}{4}
\end{array} \right.\]
\[ \Leftrightarrow - \frac{1}{4} \le x \le \frac{3}{4}\]    (thỏa mãn điều kiện xác định) (điều phải chứng minh)
b) Giải phương trình \[\sqrt {3 - 4x} + \sqrt {4x + 1} = - 16{x^2} - 8x + 1\]
Ta có:
\[VP = - \left( {16{x^2} + 8x - 1} \right)\]
\[ \Leftrightarrow VP = - \left[ {{{\left( {4x} \right)}^2} + 2.4x + 1 - 2} \right]\]
\[ \Leftrightarrow VP = - {\left( {4x + 1} \right)^2} + 2\] (1)
Ta có:
\[{\left( {4x + 1} \right)^2} \ge 0\forall x \in R\]
\[ \Leftrightarrow - {\left( {4x + 1} \right)^2} \le 0\forall x \in R\]
\[ \Leftrightarrow - {\left( {4x + 1} \right)^2} + 2 \le 2\forall x \in R\] (2)
Thay (2) vào (1), ta có:
\[VP = - {\left( {4x + 1} \right)^2} + 2 \le 2\]
Mà \[\sqrt {3 - 4x} + \sqrt {4x + 1} \ge 2\] (chứng minh câu a)
\[ \Leftrightarrow \sqrt {3 - 4x} + \sqrt {4x + 1} = - 16{x^2} - 8x + 1 = 2\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {3 - 4x} + \sqrt {4x + 1} = 2\\
- 16{x^2} - 8x + 1 = 2
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
- \left( {4x + 1} \right)\left( {4x - 3} \right) = 0\\
4x + 1 = 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
4x + 1 = 0\\
4x - 3 = 0
\end{array} \right.\\
4x + 1 = 0
\end{array} \right.\]
\[ \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = - \frac{1}{4}\\
x = \frac{3}{4}
\end{array} \right.\\
x = - \frac{1}{4}
\end{array} \right.\]
\[ \Leftrightarrow x = - \frac{1}{4}\]

Bài 07:
\[A = \left[ {\left( {\frac{1}{{\sqrt x }} + \frac{1}{{\sqrt y }}} \right)\frac{2}{{\sqrt x + \sqrt y }} + \frac{1}{x} + \frac{1}{y}} \right]:\left( {\frac{{\sqrt {{x^3}} + y\sqrt x + x\sqrt y + \sqrt {{y^3}} }}{{\sqrt {{x^3}y} + \sqrt {x{y^3}} }}} \right)\]

Điều kiện xác định:

\[ \left\{ \begin{array}{l}
x > 0\\
y > 0
\end{array} \right.\]

a) Rút gọn \[A\]
\[A = \left[ {\left( {\frac{1}{{\sqrt x }} + \frac{1}{{\sqrt y }}} \right)\frac{2}{{\sqrt x + \sqrt y }} + \frac{1}{x} + \frac{1}{y}} \right]:\left( {\frac{{\sqrt {{x^3}} + y\sqrt x + x\sqrt y + \sqrt {{y^3}} }}{{\sqrt {{x^3}y} + \sqrt {x{y^3}} }}} \right)\]
\[ \Leftrightarrow A = \left( {\frac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\frac{2}{{\sqrt x + \sqrt y }} + \frac{1}{x} + \frac{1}{y}} \right)\frac{{\sqrt {xy} \left( {x + y} \right)}}{{\left( {x + y} \right)\left( {\sqrt x + \sqrt y } \right)}}\]
\[ \Leftrightarrow A = \left( {\frac{2}{{\sqrt {xy} }} + \frac{1}{x} + \frac{1}{y}} \right)\frac{{\sqrt {xy} }}{{x + y}}\]
\[ \Leftrightarrow A = \frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{xy}}\frac{{\sqrt {xy} }}{{x + y}}\]
\[ \Leftrightarrow A = \frac{{x + 2\sqrt {xy} + y}}{{\sqrt {xy} \left( {x + y} \right)}}\]
\[ \Leftrightarrow A = \frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{\sqrt {{x^3}y} + \sqrt {x{y^3}} }}\]

Bài 01:
Cho biểu thức: \[A = \sqrt {x - 1 - 2\sqrt {x - 2} } + \sqrt {x + 7 - 6\sqrt {x - 2} } \]
a) Rút gọn \[A\]
b) Tìm \[Min\left( A \right)\]

Bài 24:
Tính \[P = \left( {1 + \frac{{\sqrt a }}{{\sqrt b }}} \right)\left( {1 + \frac{{\sqrt b }}{{\sqrt c }}} \right)\left( {1 + \frac{{\sqrt c }}{{\sqrt a }}} \right)\], biết \[a\sqrt a + b\sqrt b + c\sqrt c = 3\sqrt {abc} \]
\[P = \left( {1 + \frac{{\sqrt a }}{{\sqrt b }}} \right)\left( {1 + \frac{{\sqrt b }}{{\sqrt c }}} \right)\left( {1 + \frac{{\sqrt c }}{{\sqrt a }}} \right)\]
\[ \Leftrightarrow a,b,c \ge 0\]
Ta có:
\[a\sqrt a + b\sqrt b + c\sqrt c = 3\sqrt {abc} \]
\[ \Leftrightarrow \sqrt {{a^3}} + \sqrt {{b^3}} + \sqrt {{c^3}} - 3\sqrt {abc} = 0\]
Đặt:
\[\left\{ \begin{array}{l}
\sqrt a = x\\
\sqrt b = y\\
\sqrt c = y
\end{array} \right.\]
\[ \Leftrightarrow {x^3} + {y^3} + {z^3} - 3xyz = 0\]
\[ \Leftrightarrow \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right) = 0\]
Mà \[x + y + z > 0\]
\[ \Leftrightarrow {x^2} + {y^2} + {z^2} - xy - yz - xz = 0\]
\[ \Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {x - z} \right)^2} = 0\]
\[ \Leftrightarrow x = y = z\]
Vì:
\[\left\{ \begin{array}{l}
\sqrt a = x\\
\sqrt b = y\\
\sqrt c = y
\end{array} \right.\]
\[ \Leftrightarrow P = \left( {1 + \frac{{\sqrt a }}{{\sqrt a }}} \right)\left( {1 + \frac{{\sqrt a }}{{\sqrt a }}} \right)\left( {1 + \frac{{\sqrt a }}{{\sqrt a }}} \right)\]
\[ \Leftrightarrow P = {2^3} = 8\]
Do: 
\[\left\{ \begin{array}{l}
x = \sqrt a \ge 0\\
y = \sqrt b \ge 0\\
z = \sqrt c \ge 0
\end{array} \right.\]
\[\sqrt a ,\sqrt b ,\sqrt c \ne 0\] (vì \[\sqrt a ,\sqrt b ,\sqrt c \] là từng mẫu của P)
\[ \Leftrightarrow x + y + z > 0\]

Bài 15:
Tính tổng:
\[\frac{1}{{\sqrt 2 + 1}} + \frac{1}{{\sqrt 3 + \sqrt 2 }} + \frac{1}{{\sqrt 4 + \sqrt 3 }} + ... + \frac{1}{{\sqrt n + \sqrt {n - 1} }}\]
\[ = \frac{{\sqrt 2 - 1}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} + \frac{{\sqrt 3 - \sqrt 2 }}{{\left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)}} + \frac{{\sqrt 4 - \sqrt 3 }}{{\left( {\sqrt 4 + \sqrt 3 } \right)\left( {\sqrt 4 - \sqrt 3 } \right)}} + ... + \frac{{\sqrt n - \sqrt {n - 1} }}{{\left( {\sqrt n + \sqrt {n - 1} } \right)\left( {\sqrt n - \sqrt {n - 1} } \right)}}\]
\[ = \frac{{\sqrt 2 - 1}}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} + ... + \frac{{\sqrt n - \sqrt {n - 1} }}{{n - \left( {n - 1} \right)}}\]
\[ = \left( {\sqrt 2 - 1} \right) + \left( {\sqrt 3 - \sqrt 2 } \right) + ... + \left( {\sqrt n - \sqrt {n - 1} } \right)\]
\[ = \sqrt n - 1\]