Question

giải PT:

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

Answer 1

Lời giải:

Áp dụng BĐT AM-GM:

\(\sqrt{x^2+x-1}=\sqrt{1.(x^2+x-1)}\leq \frac{(x^2+x-1)+1}{2}\)

\(\sqrt{-x^2+x+1}=\sqrt{1(-x^2+x+1)}\leq \frac{1+(-x^2+x+1)}{2}\)

Cộng theo vế:

\(\Rightarrow x^2-x+2\leq \frac{(x^2+x-1)+1}{2}+\frac{1+(-x^2+x+1)}{2}\)

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

Mà \((x-1)^2\geq 0\forall x\in\mathbb{R}\)

Do đó \((x-1)^2=0\Leftrightarrow x=1\)