Question

Giải bpt sau : $\sqrt{x^{2}-1}$ + $\sqrt{x^{2}-x}$ $\leq$ $\sqrt{x^{2}+x-2}$

Answer 1

ĐK: \(x\ge1;x\le-2\)

\(\sqrt{x^2-1}+\sqrt{x^2-x}\le\sqrt{x^2+x-2}\)

\(\Leftrightarrow2x^2-x-1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le x^2+x-2\)

\(\Leftrightarrow x^2-2x+1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)

\(\Leftrightarrow\left(x-1\right)^2+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\left(x^2-1\right)\left(x^2-x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy bất phương trình có nghiệm \(x=1\)