Question

Giải pt \(\sqrt{x^2-x+1}=x-1\)

Answer 1

ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{x^2-x+1}=x-1\)

\(\Leftrightarrow\left|x^2-x+1\right|=\left(x-1\right)^2\)

mà \(\left|x^2-x+1\right|=x^2-x+1\forall x\)(Vì \(x^2-x+1>0\forall x\))

nên \(x^2-x+1=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-x+1=x^2-2x+1\)

\(\Leftrightarrow x^2-x+1-x^2+2x-1=0\)

\(\Leftrightarrow x=0\)(loại)

Vậy: S=\(\varnothing\)