ĐKXĐ: \(\begin{cases}2+x\ge0\\ 2-x\ge0\end{cases}=>-2\le x\le2\)
\(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\)
\(=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left\lbrack\left(\sqrt{2+x}\right)^2+\sqrt{2+x}\cdot\sqrt{2-x}+\left(\sqrt{2-x}\right)^2\right\rbrack\)
\(=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(2+x+\sqrt{4-x^2}+2-x\right)=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)\)
Ta có: \(A=\frac{\sqrt{2+\sqrt{4-x^2}}\cdot\left\lbrack\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right\rbrack}{4+\sqrt{4-x^2}}\)
\(=\frac{\sqrt{2+\sqrt{4-x^2}}\cdot\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)}{4+\sqrt{4-x^2}}=\sqrt{2+\sqrt{4-x^2}}\cdot\left(\sqrt{2+x}-\sqrt{2-x}\right)\)
Đặt \(B=\sqrt{2+x}-\sqrt{2-x}\)
=>\(B^2=2+x+2-x-2\cdot\sqrt{\left(2+x\right)\left(2-x\right)}=4-2\cdot\sqrt{4-x^2}\)
=>\(B=\sqrt{4-2\cdot\sqrt{4-x^2}}\)
TA có: \(A=\sqrt{2+\sqrt{4-x^2}}\cdot\left(\sqrt{2+x}-\sqrt{2-x}\right)\)
\(=\sqrt{2+\sqrt{4-x^2}}\cdot\sqrt{4-2\cdot\sqrt{4-x^2}}\)
\(=\sqrt{2+\sqrt{4-x^2}}\cdot\sqrt2\cdot\sqrt{2-\sqrt{4-x^2}}\)
\(=\sqrt2\cdot\sqrt{4-\left(4-x^2\right)}=\sqrt2\cdot\sqrt{x^2}=\sqrt2\cdot\left|x\right|\)
TH1: 0<=x<=2
=>2+x>=2-x
=>\(\sqrt{x+2}\ge\sqrt{2-x}\)
=>\(\sqrt{x+2}-\sqrt{2-x}\ge0\)
=>A>=0
=>\(A=\sqrt2\cdot x\) (2)
TH2: -2<=x<0
=>2-x>2+x
=>\(\sqrt{2-x}-\sqrt{2+x}>0\)
=>\(\sqrt{2+x}-\sqrt{2-x}<0\)
=>A<0
=>\(A=-\left|x\right|\cdot\sqrt2=-\left(-x\right)\cdot\sqrt2=x\sqrt2\) (1)
Từ (1),(2) suy ra \(A=x\sqrt2\)
