Question

giải pt

\(\sqrt{\left(x-3\right)^2}=3-x\)

 

Answer 1

`ĐKXĐ: x <= 3`

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

`<=>|x-3|=3-x`

`TH1: x-3 = 3-x`

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

`<=>2x=6`

`<=>x=3(tm)`

`TH2: x-3 = -(3-x)`

`<=>x-3=x-3`(luôn đúng `forall x <= 3`)

Vậy phương trình có tập nghiệm `x <= 3`

Answer 2

\(\sqrt{\left(x-3\right)^2}=3-x\) `ĐKXĐ : x >= 3`

\(\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=3-x\\x-3=-\left(3-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3-3+x=0\\x-3=-3+x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\x-3=x-3\left(luôn-đúng\right)\end{matrix}\right.\)

`=> 2x - 6 =0`

`=>x = 3 (tm) `