Question

[Minh Bao]

(8x-4)(-x²+2x-2)=0

Answer 1

[Nguyễn Ngọc Anh Minh]

Ta nhận thấy

\(-x^2+2x-2=-\left[\left(x^2-2x+1\right)+1\right]\)

Ta có

\(x^2-2x+1\ge0\Rightarrow\left(x^2-2x+1\right)+1\ge1\)

\(\Rightarrow-\left[\left(x^2-2x+1\right)+1\right]\le-1\)

\(\Rightarrow PT\Leftrightarrow8x-4=0\Leftrightarrow x=\dfrac{1}{2}\)

Answer 2

[Nguyễn Huy Tú]

\(\left(8x-4\right)\left(-x^2+2x-2\right)=0\Leftrightarrow\left(8x-4\right)\left(x^2-2x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}8x-4=0\\x^2-2x+2=0\left(loai\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)

do \(x^2-2x+2=x^2-2x+1+1=\left(x-1\right)^2+1>0\)