Question

Tìm x biết

\(\sqrt{2-x}+\sqrt{2+x}+\sqrt{4-x^2}=2\)

Answer 1

đk: \(-2\le x\le2\)

pt <=> \(\sqrt{2-x}+\sqrt{2+x}=2-\sqrt{4-x^2}\)

bình phương 2 vế:\(\Leftrightarrow2-x+2+x+2\sqrt{4-x^2}=4+4-x^2-4\sqrt{4-x^2}\)

\(\Leftrightarrow4-x^2-6\sqrt{4-x^2}=0\) (*)

Đặt \(\sqrt{4-x^2}=a\left(0\le a\le2\right)\)

(*) <=> \(a^2-6a=0\Leftrightarrow a\left(a-6\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\left(tm\right)\\a=6\left(ktm\right)\end{matrix}\right.\)

a = 0 <=> 4 -x² = 0 <=> x² = 4 \(\Leftrightarrow x=\pm2\)

vậy...