Đáp án:

Giải thích các bước giải:

`2.`

`\sqrt{x^2-9}-\sqrt{x-3}=0` `(đk:`$\left[\begin{matrix} x\le-3\\ x\ge3\end{matrix}\right.$`)`

`<=>\sqrt{x^2-9}=\sqrt{x-3}`

`<=>x^2-9=x-3`

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

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

`<=>(x-3)(x+2)=0`

`<=>`$\left[\begin{matrix} x=3(n)\\ x=-2(l)\end{matrix}\right.$

`S={3}`

`3.`

`\sqrt{x^2-25}-\sqrt{x+5}=0` `(đk:`$\left[\begin{matrix} x\le-5\\ x\ge5\end{matrix}\right.$`)`

`<=>\sqrt{x^2-25}=\sqrt{x+5}`

`<=>x^2-25=x+5`

`<=>(x-5)(x+5)-(x+5)=0`

`<=>(x+5)(x-5-1)=0`

`<=>(x+5)(x-6)=0`

`<=>`$\left[\begin{matrix} x=-5(n)\\ x=6(n)\end{matrix}\right.$

`S={-5;6}`

`4.`

`\sqrt{x^2-3x+2}-\sqrt{x-1}=0` `(đk:`$\left[\begin{matrix} x\le1\\ x\ge2\end{matrix}\right.$`)`

`<=>\sqrt{x^2-3x+2}=\sqrt{x-1}`

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

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

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

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

`<=>`$\left[\begin{matrix} x=1(n)\\ x=3(n)\end{matrix}\right.$

`S={1;3}`

`#Devil`

Đáp án:

$\begin{array}{l}
2)Dkxd:x \ge 3\\
\sqrt {{x^2} - 9}  - \sqrt {x - 3}  = 0\\
 \Leftrightarrow \sqrt {x - 3} .\sqrt {x + 3}  - \sqrt {x - 3}  = 0\\
 \Leftrightarrow \sqrt {x - 3} .\left( {\sqrt {x + 3}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
\sqrt {x + 3}  = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x + 3 = 1 \Leftrightarrow x =  - 2\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 3\\
3)DKxd:\left\{ \begin{array}{l}
{x^2} - 25\\
x \ge  - 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 5\\
x \le  - 5
\end{array} \right.\\
x \ge  - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 5\\
x =  - 5
\end{array} \right.\\
\sqrt {{x^2} - 25}  - \sqrt {x + 5}  = 0\\
 \Leftrightarrow \sqrt {x - 5} .\sqrt {x + 5}  - \sqrt {x + 5}  = 0\\
 \Leftrightarrow \sqrt {x + 5} \left( {\sqrt {x - 5}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 5\left( {tm} \right)\\
x - 5 = 1 \Leftrightarrow x = 6\left( {tm} \right)
\end{array} \right.\\
Vậy\,x =  - 5;x = 6\\
3)Dkxd:\left\{ \begin{array}{l}
{x^2} - 3x + 2 \ge 0\\
x - 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 1} \right) \ge 0\\
x \ge 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
x \ge 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x = 1
\end{array} \right.\\
\sqrt {{x^2} - 3x + 2}  - \sqrt {x - 1}  = 0\\
 \Leftrightarrow \sqrt {x - 2} .\sqrt {x - 1}  - \sqrt {x - 1}  = 0\\
 \Leftrightarrow \sqrt {x - 1} \left( {\sqrt {x - 2}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
\sqrt {x - 2}  = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x - 2 = 1 \Leftrightarrow x = 3\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 1;x = 3
\end{array}$