`Parkchaeyoung`

`a,`

$\sqrt{x^2-6x+9}$ `=4-x`

`<=>` $\sqrt{(x-3)^2}$ `=4-x`

`<=> |x-3|=4-x` `(1)`

`TH1: x-3>=0=>x>=3=>|x-3|=x-3`

`(1)=> x-3=4-x`

`<=> x+x=4+3`

`<=> 2x=7`

`<=> x=7/2` `(tm)`

`TH2: x-3<0=>x<3=>|x-3|=-x+3`

`(1)=> -x+3=4-x`

`<=> -x+x=4-3`

`<=> 0x=1` (vô lí)

Vậy `x=7/2`

`b,` 

$\sqrt{4x^2-4x+1}$ `=5x-3`

`<=>` $\sqrt{(2x-1)^2}$ `=5x-3`

`<=> |2x-1|=5x-3` `(2)`

`TH1: 2x-1>=0=>x>=1/2=>|2x-1|=2x-1`

`(2)=> 2x-1=5x-3`

`<=> 2x-5x=-3+1`

`<=> -3x=-2`

`<=> x=2/3` `(tm)`

`TH2: 2x-1<0=>x<1/2=>|2x-1|=-2x+1`

`(2)=> -2x+1=5x-3`

`<=> -2x-5x=-3-1`

`<=> -7x=-4`

`<=> x=4/7` (ko tm)

Vậy `x=2/3`

Đáp án:`+`Giải thích các bước giải:

`a,`

`\sqrt{x^2-6x+9}=4-x`

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

`<=>|x-3|=4-x`   `ĐK:x<4`

`<=>` $\left[\begin{matrix} x-3=4-x\\ x-3=-4+x\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x+x=4+3\\ x-x=-4+3\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x=7\\ 0x=-1(vô lí)\end{matrix}\right.$

`<=>` `x=7/2(tm)`

Vậy `S={7/2}`

`b,`

`\sqrt{4x^2-4x+1}=5x-3`

`<=>(sqrt{2x-1})^2=5x-3`

`<=>|2x-1|=5x-3`   `ĐK:x>=3/5`

`<=>` $\left[\begin{matrix} 2x-1=5x-3\\ 2x-1=-5x+3\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x-5x=-3+1\\ 2x+5x=3+1\end{matrix}\right.$

`<=>` $\left[\begin{matrix} -3x=-2\\ 7x=4\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=2/3(ktm)\\ x=4/7(tm)\end{matrix}\right.$

Vậy `S={4/7}`