Question

\(x+\sqrt{5-x^2}+x\sqrt{5-x^2}=5\)

Answer 1

\(x+\sqrt{5-x^2}+x\sqrt{5-x^2}=5\)

ĐKXĐ: \(-\sqrt{5}\le x\le\sqrt{5}\)

Đặt \(\sqrt{5-x^2}=t\)(\(t\)\(\ge0\))

Ta có: \(\left\{{}\begin{matrix}x+t+xt=5\\x^2+t^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x+t\right)+2xt=10\\\left(x+t\right)^2-2xt=5\end{matrix}\right.\)

\(\Rightarrow\left(x+t\right)^2+2\left(x+t\right)-15=0\Leftrightarrow\left(x+t+5\right)\left(x+t-3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=-5-x\\t=3-x\end{matrix}\right.\)(giải hai trường hợp rồi kết luận nghiệm)