Đáp án:
$a) A=\frac{-3}{2\sqrt{x}(\sqrt{x}-3)}\\
b)
B(12+6\sqrt{3})=\frac{-1}{2\sqrt{3}+2}$
Giải thích các bước giải:
$a) A=\left ( \frac{x}{\sqrt{x}+3} -\frac{x+1}{\sqrt{x}-3}+\frac{6x+\sqrt{x}}{x-9}\right ):\left ( \frac{\sqrt{x}-3}{\sqrt{x}+3}+1 \right )\\
=\left ( \frac{x(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)} -\frac{(\sqrt{x}+3)(x+1)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\frac{6x+\sqrt{x}}{x-9}\right ):\left ( \frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}+3} \right )\\
=\frac{x\sqrt{x}-3x-x\sqrt{x}-\sqrt{x}-3x-3+6x+\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}:(\frac{\sqrt{x}-3+\sqrt{x}+3}{\sqrt{x}+3})\\
=\frac{-3}{(\sqrt{x}+3)(\sqrt{x}-3)}:\frac{2\sqrt{x}}{\sqrt{x}+3}\\
=\frac{-3}{(\sqrt{x}+3)(\sqrt{x}-3)}.\frac{\sqrt{x}+3}{2\sqrt{x}}\\
=\frac{-3}{2\sqrt{x}(\sqrt{x}-3)}\\
b)12+6\sqrt{3}=3^2+2.3.\sqrt{3}+3=(3+\sqrt{3})^2\\
B(12+6\sqrt{3})=\frac{-3}{2\sqrt{(3+\sqrt{3})^2}(\sqrt{(3+\sqrt{3})^2}-3)}\\
=\frac{-3}{2(3+\sqrt{3})(3+\sqrt{3}-3)}\\
=\frac{-3}{2(3+\sqrt{3}).\sqrt{3}}\\
=\frac{-3}{6\sqrt{3}+6}\\
=\frac{-1}{2\sqrt{3}+2}$
