Question

sqrt(x ^ 2 - 2x) = 2 - x Giúp e ạ

Answer 1

ĐK: $\begin{cases}x^2-2x≥0\\2-x≥0\\\end{cases}$ `<=>` $\begin{cases}x≤0 \vee x ≥2\\x≤2\\\end{cases}$ ` <=> ` $\begin{cases}x≤0\\x=2\\\end{cases}$

`\sqrt(x^2-2x)=2-x`

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

`<=>x^2-2x=x^2-4x+4`

`<=>-2x=-4x+4`

`<=>x=2`

Vậy `S={2}`.

 

Answer 2

Ta có: \(\sqrt{x^2-2x}=2-x\)

\(\Leftrightarrow x^2-2x=x^2-4x+4\)

\(\Leftrightarrow-2x+4x=4\)

\(\Leftrightarrow2x=4\)

hay x=2