Question

giả pt vô tỉ \(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)

Answer 1

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow\frac{x^2+x-2-2x+2}{\sqrt{x^2+x-2}+\sqrt{2x-2}}+\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\frac{x\left(x-1\right)}{\sqrt{x^2+x-2}+\sqrt{2x-2}}+\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x}{\sqrt{x^2+x-2}+\sqrt{2x-2}}+x+1\right)=0\)

\(\Leftrightarrow x-1=0\) (ngoặc phía sau luôn dương \(\forall x>1\))

\(\Rightarrow x=1\)