\(ĐK:-2\le x\le2\)
Đặt \(x+\sqrt{4-x^2}=a\ge0\Leftrightarrow2x\sqrt{4-x^2}=a^2-4\)
\(PT\Leftrightarrow\dfrac{3}{2}\left(a^2-4\right)+2=a\\ \Leftrightarrow3a^2-12+4=2a\\ \Leftrightarrow3a^2-2a-8=0\\ \Leftrightarrow a=2\left(a\ge0\right)\\ \Leftrightarrow2x\sqrt{4-x^2}=4-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\pm2\left(tm\right)\end{matrix}\right.\)
ĐKXĐ: \(-2\le x\le2\)
Đặt \(x+\sqrt{4-x^2}=a\)
\(\Rightarrow a^2=4+2x\sqrt{4-x^2}\)
\(\Rightarrow x\sqrt{4-x^2}=\dfrac{a^2-4}{2}\) (1)
Phương trình trở thành:
\(a=2+\dfrac{3\left(a^2-4\right)}{2}\)
\(\Leftrightarrow3a^2-2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\sqrt{4-x^2}=2\\x+\sqrt{4-x^2}=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4-x^2}=2-x\\\sqrt{4-x^2}=-\dfrac{4}{3}-x\left(x\le-\dfrac{4}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x^2=x^2-4x+4\\4-x^2=x^2+\dfrac{8}{3}x+\dfrac{16}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=0\\2x^2+\dfrac{8}{3}x-\dfrac{20}{9}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{-2+\sqrt{14}}{3}\left(loại\right)\\x=\dfrac{-2-\sqrt{14}}{3}\end{matrix}\right.\)
