Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
DKXD:\,\,\,x + 1 \ge 0 \Leftrightarrow x \ge  - 1\\
\sqrt {x + 1}  = x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
x + 1 = {\left( {x - 1} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x + 1 = {x^2} - 2x + 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
{x^2} - 3x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x\left( {x - 3} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 3\\
2,\\
DKXD:\,\,\,\,2x + 3 \ge 0 \Leftrightarrow x \ge  - \frac{3}{2}\\
x - \sqrt {2x + 3}  = 0\\
 \Leftrightarrow x = \sqrt {2x + 3} \\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} = 2x + 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
{x^2} - 2x - 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left( {x - 3} \right)\left( {x + 1} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left[ \begin{array}{l}
x = 3\\
x =  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 3\\
3,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x + 4 \ge 0\\
1 - x \ge 0\\
1 - 2x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 4\\
x \le 1\\
x \le \frac{1}{2}
\end{array} \right. \Leftrightarrow  - 4 \le x \le \frac{1}{2}\\
\sqrt {x + 4}  - \sqrt {1 - x}  = \sqrt {1 - 2x} \\
 \Leftrightarrow \sqrt {x + 4}  = \sqrt {1 - x}  + \sqrt {1 - 2x} \\
 \Leftrightarrow x + 4 = {\left( {\sqrt {1 - x}  + \sqrt {1 - 2x} } \right)^2}\\
 \Leftrightarrow x + 4 = \left( {1 - x} \right) + 2.\sqrt {1 - x} .\sqrt {1 - 2x}  + \left( {1 - 2x} \right)\\
 \Leftrightarrow x + 4 = 2 - 3x + 2\sqrt {\left( {1 - x} \right)\left( {1 - 2x} \right)} \\
 \Leftrightarrow 4x + 2 = 2\sqrt {1 - 3x + 2{x^2}} \\
 \Leftrightarrow 2x + 1 = \sqrt {2{x^2} - 3x + 1} \\
 \Leftrightarrow \left\{ \begin{array}{l}
2x + 1 \ge 0\\
4{x^2} + 4x + 1 = 2{x^2} - 3x + 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \frac{1}{2}\\
2{x^2} + 7x = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \frac{1}{2}\\
\left[ \begin{array}{l}
x = 0\\
x =  - \frac{7}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 0\,\,\,\,\,\left( {t/m} \right)\\
4,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x - 2 \ge 0\\
{x^2} - 4 \ge 0
\end{array} \right. \Leftrightarrow x \ge 2\\
\sqrt {x - 2}  - 3\sqrt {{x^2} - 4}  = 0\\
 \Leftrightarrow \sqrt {x - 2}  - 3.\sqrt {\left( {x - 2} \right)\left( {x + 2} \right)}  = 0\\
 \Leftrightarrow \sqrt {x - 2} .\left( {1 - 3\sqrt {x + 2} } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2}  = 0\\
1 - 3\sqrt {x + 2}  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
\sqrt {x + 2}  = \frac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x + 2 = \frac{1}{9}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \frac{{17}}{9}
\end{array} \right.\\
x \ge 2 \Rightarrow x = 2
\end{array}\)

Đáp án:

1. `sqrt(x+1)=x-1` (Đk: `x>= 1`)

`<=> (sqrt(x+1))^2=(x-1)^2`

`<=> x+1=x^2-2x+1`

`<=> x+1-x^2+2x-1=0`

`<=> -x^2+3x=0`

`<=> -x(x-3)=0`

TH1:

`x=0` (KTMĐK)

TH2:

`x-3=0`

`<=> x=3` (t/m)

Vậy `S={3}`

2. `x-sqrt(2x+3)=0` (Đk: `x>= -3/2`)

`<=> x=sqrt(2x+3)`

`<=> x^2=2x+3`

`<=> x^2-2x-3=0`

`<=> x^2+x-3x-3=0`

`<=> x(x+1)-3(x+1)=0`

`<=> (x+1)(x-3)=0`

TH1:

`x+1=0`

`<=> x=-1` (KTMĐK)

TH2:

`x-3=0`

`<=> x=3` (t/m)

Vậy `S={3}`

3. `sqrt(x+4)-sqrt(1-x)=sqrt(1-2x)` (Đk: `-4<=x<= 1/2`)

`<=> sqrt(x+4)=sqrt(1-2x)+sqrt(1-x)`

`<=> (sqrt(x+4))^2=(sqrt(1-2x)+sqrt(1-x))^2`

`<=> x+4=(sqrt(1-2x))^2+2sqrt(1-2x)*sqrt(1-x)+(sqrt(1-x))^2`

`<=> x+4=1-2x+2sqrt((1-2x)*(1-x))+1-x`

`<=> x+4=(1-2x+1-x)+2sqrt(1-3x+2x^2)`

`<=> x+4=2-3x+2sqrt(1-3x+2x^2)`

`<=> x+4-2+3x=2sqrt(1-3x+2x^2)`

`<=> 4x+2=2sqrt(1-3x+2x^2)`

`<=> (4x+2)^2=(2sqrt(1-3x+2x^2))^2`

`<=> 16x^2+16x+4=4(1-3x+2x^2)`

`<=> 16x^2+16x+4=4-12x+8x^2`

`<=> 16x^2+16x+4-4+12x-8x^2=0`

`<=> 8x^2+28x=0`

`<=> 4x(2x+7)=0`

TH1:

`4x=0`

`<=> x=0` (t/m)

TH2:

`2x+7=0`

`<=> 2x=-7`

`<=> x=-7/2` (KTMĐK)

Vậy `S={0}`

4. `sqrt(x-2)-3sqrt(x^2-4)=0` (Đk: `x>=2`)

`<=> sqrt(x-2)=3sqrt(x^2-4)`

`<=> (sqrt(x-2))^2=(3sqrt(x^2-4))^2`

`<=> x-2=9(x^2-4)`

`<=> (x-2)-9(x^2-2^2)=0`

`<=> (x-2)-9(x+2)(x-2)=0`

`<=> (x-2)[1-9(x+2)]=0`

`<=> (x-2)(1-9x-18)=0`

`<=> (x-2)(-17-9x)=0`

TH1:

`x-2=0`

`<=> x=2` (t/m)

TH2:

`-17-9x=0`

`<=> 9x=-17`

`<=> x=-17/9` (KTMĐK)

Vậy `S={2}`