\(\sqrt{4-\sqrt{4+x}}=x\left(Đkxđ:x\ge-4\right)\)
\(\Leftrightarrow4-\sqrt{4+x}=x^2\)
\(\Leftrightarrow4-x^2=\sqrt{4+x}\)
\(\Leftrightarrow\left(4-x^2\right)^2=4+x\left(đkxđ:x^2\le4\right)\)
\(\Leftrightarrow16-8x^2+x^4=4+x\left(-2\le x\le2\right)\)
\(\Leftrightarrow x^4-8x^2-x+12=0\)
\(\Leftrightarrow\left(x^2-x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-4=0\\x^2+x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{17}}{2}\\x=\frac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
Từ: \(Đkxđ:-2\le x\le2\) ta có:
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1\sqrt{17}}{2}\\x=\frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
Vậy ............
