Question

Cho \(\sqrt{x^2-4x+8}=2\)

Answer 1

ĐK: `x^2-4x+8>=0 <=> (x^2-4x+4)+4>=0 <=>(x-2)^2+4>=0 forall x`

`\sqrt(x^2-4x+8)=2`

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

`<=>x^2-4x+4=0`

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

`<=>x=2`

Answer 2

Ta có: \(\sqrt{x^2-4x+8}=2\)

\(\Leftrightarrow x^2-4x+8=4\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow x-2=0\)

hay x=2