Question

\(\sqrt{25-x^2}-\sqrt{10-x^2}=3\)

Answer 1

ĐKXĐ: \(-\sqrt{10}\le x\le\sqrt{10}\)

Ta có: \(\sqrt{25-x^2}-\sqrt{10-x^2}=3\)

\(\Leftrightarrow\sqrt{25-x^2}=3+\sqrt{10-x^2}\)

\(\Leftrightarrow\sqrt{\left(25-x^2\right)^2}=\left(3+\sqrt{10-x^2}\right)^2\)

\(\Leftrightarrow25-x^2=9+6\cdot\sqrt{10-x^2}+10-x^2\)

\(\Leftrightarrow25-x^2=6\sqrt{10-x^2}-x^2+19\)

\(\Leftrightarrow25-x^2-6\sqrt{10-x^2}+x^2-19=0\)

\(\Leftrightarrow-6\sqrt{10-x^2}+6=0\)

\(\Leftrightarrow-6\sqrt{10-x^2}=-6\)

\(\Leftrightarrow10-x^2=1\)

\(\Leftrightarrow x^2=9\)

hay \(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)

Vậy: S={3;-3}