ĐKXĐ: \(\begin{cases}9-x^2\ge0\\ 1+\frac{2x}{1-x}>0\end{cases}\Rightarrow\begin{cases}x^2\le9\\ \frac{1-x+2x}{1-x}>0\end{cases}\)
=>\(\begin{cases}-3\le x<=3\\ \frac{1+x}{1-x}>0\end{cases}\Rightarrow\begin{cases}-3\le x\le3\\ \frac{x+1}{x-1}<0\end{cases}\)
=>-3<=x<=3 và -1<x<1
=>-1<x<1
Ta có: \(A=\frac{x^2+5x+6+x\cdot\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\cdot\sqrt{9-x^2}}:2\sqrt{1+\frac{2x}{1-x}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\cdot\sqrt{3-x}\cdot\sqrt{3+x}}{x\left(3-x\right)+\left(x+2\right)\cdot\sqrt{3-x}\cdot\sqrt{3+x}}:2\sqrt{\frac{1-x+2x}{1-x}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\cdot\sqrt{3+x}+x\cdot\sqrt{3-x}\right)}{\sqrt{3-x}\left(x\cdot\sqrt{3-x}+\left(x+2\right)\cdot\sqrt{3+x}\right)}:2\sqrt{\frac{1+x}{1-x}}\)
\(=\sqrt{\frac{3+x}{3-x}}:\sqrt{\frac{4+4x}{1-x}}=\sqrt{\frac{3+x}{3-x}:\frac{4\left(x+1\right)}{1-x}}=\sqrt{\frac{3+x}{3-x}\cdot\frac{1-x}{4\left(x+1\right)}}\)
\(=\sqrt{\frac{\left(3+x\right)\left(1-x\right)}{4\left(3-x\right)\left(x+1\right)}}\)
