Question

Rút gọn biểu thức: 

Answer 1

Đk: \(-3\le x\le3\) và mẫu khác 0

\(A=\dfrac{x^2+5x+x\sqrt{9-x^2}+6}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)\(=\dfrac{\left(x+3\right)\left(x+2\right)+x\sqrt{\left(x+3\right)\left(3-x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(x+3\right)\left(3-x\right)}}\)\(=\dfrac{\sqrt{x+3}\left[\sqrt{x+3}\left(x+2\right)+x\sqrt{3-x}\right]}{\sqrt{3-x}\left[\sqrt{x+3}\left(x+2\right)+x\sqrt{3-x}\right]}\)\(=\dfrac{\sqrt{9-x^2}}{3-x}\)