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\[\begin{array}{l}
\Leftrightarrow x - 1 + \frac{{x - 1}}{{\sqrt {2x + 1} + \sqrt {x + 2} }} = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {1 + \frac{1}{{\sqrt {2x + 1} + \sqrt {x + 2} }}} \right) = 0
\end{array}\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
1 + \frac{1}{{\sqrt {2x + 1} + \sqrt {x + 2} }} = 0
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x = 1 (Nhận) \\
\sqrt {2x + 1} + \sqrt {x + 2} = - 1
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x = 1  (Nhận) \\
2\sqrt {(2x + 1)(x + 2)} = - 3x - 2
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x = 1  (Nhận) \\
4(2x + 1)(x + 2) = {\left( { - 3x - 2} \right)^2}
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x = 1 (Nhận)  \\
- {x^2} + 8x + 4 = 0
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x = 1 (Nhận) \\
\left[ \begin{array}{l}
x = 2\left( {2 - \sqrt 5 } \right) (Loại)\\
x = 2\left( {2 + \sqrt 5 } \right)(Loại)
\end{array} \right.
\end{array} \right.\]
Vậy PT có 1 nghiệm duy nhất \[x = 1\]

\[\begin{array}{l}
\Leftrightarrow ({x^2} - 2x - 1)(x + 1) - \left( {\sqrt {{x^2} - 2x + 3} + 2} \right)\left( {{x^2} - 2x - 1} \right) = 0\\
\Leftrightarrow ({x^2} - 2x - 1)\left( {x - 1 - \sqrt {{x^2} - 2x + 3} } \right) = 0
\end{array}\]

\[ \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x - 1 = 0\\
\sqrt {{x^2} - 2x + 3} = x - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left[ \begin{array}{l}
x = 1 + \sqrt 2 (Nhận) \\
x = 1 - \sqrt 2 (Nhận)
\end{array} \right.\\
{x^2} - 2x + 3 = {x^2} - 2x + 1
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
\left[ \begin{array}{l}
x = 1 + \sqrt 2 (Nhận) \\
x = 1 - \sqrt 2 (Nhận)
\end{array} \right.\\
3 = 1(vô lý) \Rightarrow vô nghiệm
\end{array} \right.\]

BÀI 7:
\[\begin{array}{l}                                                             
(x + 1)\sqrt {{x^2} - 2x + 3} = {x^2} + 1\\      
\Leftrightarrow (x + 1)\sqrt {{x^2} - 2x + 3} - {x^2} - 1 = 0\\
\Leftrightarrow (x + 1)\sqrt {{x^2} - 2x + 3} - (x + 1)(x - 1) = 0\\
\Leftrightarrow (x + 1)\left[ {\sqrt {{x^2} - 2x + 3} - (x - 1)} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
\sqrt {{x^2} - 2x + 3} - x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
\sqrt {{x^2} - 2x + 3} = x - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
\left\{ \begin{array}{l}
x - 1 \ge 0\\
{x^2} - 2x + 3 = {x^2} - 2x + 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
\left\{ \begin{array}{l}
x \ge 1\\
3 = 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x \in \emptyset
\end{array} \right. \Leftrightarrow x = - 1
\end{array}\]

\[\begin{array}{l}
\sqrt {x - 1} + 1 - \sqrt {x - 1} + 1 = 2\\
\Leftrightarrow 2 = 2 (Đúng)
\end{array}\]
\[ \Leftrightarrow x \in R\]
\[*x\langle 2\]

\[\begin{array}{l}
\sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 = 2\\
\Leftrightarrow 2\sqrt {x - 1} = 2
\end{array}\]

\[\begin{array}{l}
\Leftrightarrow \sqrt {x - 1} = 1\\
\Leftrightarrow x - 1 = 1\\
\Leftrightarrow x = 2 (Loại)
\end{array}\]
\[ \Leftrightarrow x \in \emptyset \]
Vậy tập nghiệm \[S = \left\{ {x \in \left. R \right|x \ge 2} \right\}\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 3x + 1 \le 0\\
2x - 1 = {x^4} - 6{x^3} + 11{x^2} - 6x + 1
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
\frac{{3 - \sqrt 5 }}{2} \le x \le \frac{{3 + \sqrt 5 }}{2}\\
{x^4} - 6{x^3} + 11{x^2} - 8x + 2 = 0
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
\frac{{3 - \sqrt 5 }}{2} \le x \le \frac{{3 + \sqrt 5 }}{2}\\
{x^4} - 4{x^3} + 2{x^2} - 2{x^3} + 8{x^2} - 4x + {x^2} - 4x + 2 = 0
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
\frac{{3 - \sqrt 5 }}{2} \le x \le \frac{{3 + \sqrt 5 }}{2}\\
{(x - 1)^2}\left( {{x^2} - 4x + 2} \right) = 0
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
\frac{{3 - \sqrt 5 }}{2} \le x \le \frac{{3 + \sqrt 5 }}{2}\\
\left[ \begin{array}{l}
{(x - 1)^2} = 0\\
{x^2} - 4x + 2 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{{3 - \sqrt 5 }}{2} \le x \le \frac{{3 + \sqrt 5 }}{2}\\
\left[ \begin{array}{l}
x = 1 (Nhận) \\
\left[ \begin{array}{l}
x = 2 + \sqrt 2 (Loại) \\
x = 2 - \sqrt 2 (Nhận)
\end{array} \right.
\end{array} \right.
\end{array} \right.\]\[ \Leftrightarrow \] \[x = 1;x = 2 - \sqrt 2 \]
Vậy tập nghiệm \[S = \left\{ {1,2 - 2\sqrt 2 } \right\}\]

\[\begin{array}{l}
\Rightarrow a + b - 2\sqrt {ab} \le a - b\\
\Leftrightarrow b \le \sqrt {ab}
\end{array}\]

\[\begin{array}{l}
\Leftrightarrow \sqrt b \le \sqrt a \\
\Leftrightarrow \sqrt {10x + 1} - \sqrt {9x + 4} = \sqrt {2x - 2} - \sqrt {3x - 5}
\end{array}\]
Ta có: \[\sqrt {10x + 1}  \ge 0,\sqrt {9x + 4}  \ge 0,\sqrt {2x - 2}  \ge 0,\sqrt {3x - 5}  \ge 0\]
\[ * \sqrt {10x + 1}  \ge \sqrt {9x + 4} \]

\[ \Leftrightarrow \left\{ \begin{array}{l}
10x + 1 \ge 0\\
9x + 4 \ge 0\\
10x + 1 \ge 9x + 4
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{{ - 1}}{{10}}\\
x \ge \frac{{ - 4}}{9}\\
x \ge 3
\end{array} \right.\] 
\[ \Leftrightarrow \sqrt {10x + 1}  \ge \sqrt {9x + 4} (\forall x \ge 3)\]

\[\begin{array}{l}
* \sqrt {2x - 2} \ge \sqrt {3x - 5} \\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 2 \ge 0\\
3x - 5 \ge 0\\
2x - 2 \ge 3x - 5
\end{array} \right.
\end{array}\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \ge \frac{5}{3}\\
x \ge 3
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
x - 3 \ge 0\\
3 - x \ge 0\\
x - 3 = 3 - x
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
x \le 3\\
x = 3
\end{array} \right. \Leftrightarrow x = 3\]
Vậy khi \[x = 3\] thì \[\sqrt {x - 3}  = \sqrt {3 - x} \]\[ \Rightarrow \]\[\sqrt {10x + 1}  - \sqrt {9x + 4}  = \sqrt {2x - 2}  - \sqrt {3x - 5} \] (TCBC)
Vậy PT có 1 nghiệm duy nhất là \[x = 3\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
2x + 2 \ge 0\\
3x + 1 \ge 0\\
2x + 2 = 3x + 1
\end{array} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \ge \frac{{ - 1}}{3}\\
x = 1 (Nhận)
\end{array} \right.\]
\[ \Leftrightarrow x = 1\]  
Vậy khi \[x = 1\] thì \[2\sqrt {2x + 2}  = 2\sqrt {3x + 1} \]\[ \Rightarrow \sqrt {x + 3}  + \sqrt {3x + 1}  = 2\sqrt x  + \sqrt {2x + 2} \] (TCBC)

Vậy PT có 1 nghiệm duy nhất \[x = 1\]

\[ \Leftrightarrow \left[ \begin{array}{l}
t + 1 = 0\\
t - 2 = 0\\
{t^2} - 2t - 1 = 0
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
t = - 1  ( Loại) \\
t = 2 (Nhận) \\
t = 1 + \sqrt 2 (Nhận) \\
t = 1 - \sqrt 2 (Loại)
\end{array} \right.\]
Với  

\[\begin{array}{l}
* t = 2\\
\Leftrightarrow \sqrt {x + 1} = 2\\
\Leftrightarrow x + 1 = 4\\
\Leftrightarrow x = 3  (Nhận)
\end{array}\]

\[\begin{array}{l}
* t = 1 + \sqrt 2 \\
\Leftrightarrow \sqrt {x + 1} = 1 + \sqrt 2 \\
\Leftrightarrow x + 1 = 3 + 2\sqrt 2 \\
\Leftrightarrow x = 2 + 2\sqrt 2 (Nhận)
\end{array}\]
Vậy tập nghiệm \[S = \left\{ {3,2 + 2\sqrt 2 } \right\}\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
1 + \frac{{ - x + 1}}{{1 + \sqrt {{x^2} - {{\left( {2\sqrt {x - 1} } \right)}^{^2}}} }} = 0
\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
2 - x + \sqrt {{x^2} - {{\left( {2\sqrt {x - 1} } \right)}^{^2}}} \Rightarrow {x^2} - 4x + 4 = {x^2} - 4x + 4
\end{array} \right.\]   (Đúng)\[ \Leftrightarrow \]\[{x \in N}\]