Question

Bài 1: Rút gọn:

a) \(\left(\frac{\sqrt{x}+2}{x-9}-\frac{\sqrt{x}-2}{x+6\sqrt{x}+9}\right)\left(\sqrt{x}-\frac{9}{\sqrt{x}}\right)\) (x > 0, x ≠ 9)

b) \(\frac{3}{2\sqrt{3}+3}+\frac{3}{2\sqrt{3}-3}\)

(mink đag cần gấp)

Answer 1

a) Ta có: \(\left(\frac{\sqrt{x}+2}{x-9}-\frac{\sqrt{x}-2}{x+6\sqrt{x}+9}\right)\left(\sqrt{x}-\frac{9}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+3\right)^2}\right)\left(\frac{x}{\sqrt{x}}-\frac{9}{\sqrt{x}}\right)\)

\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}\right)\cdot\frac{x-9}{\sqrt{x}}\)

\(=\frac{x+5\sqrt{x}+6-\left(x-5\sqrt{x}+6\right)}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)^2}\cdot\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\frac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\sqrt{x}+3}\cdot\frac{1}{\sqrt{x}}\)

\(=\frac{10\sqrt{x}}{\sqrt{x}+3}\cdot\frac{1}{\sqrt{x}}\)

\(=\frac{10}{\sqrt{x}+3}\)

b) Ta có: \(\frac{3}{2\sqrt{3}+3}+\frac{3}{2\sqrt{3}-3}\)

\(=\frac{3\left(2\sqrt{3}-3\right)}{\left(2\sqrt{3}+3\right)\left(2\sqrt{3}-3\right)}+\frac{3\left(2\sqrt{3}+3\right)}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}\)

\(=\frac{6\sqrt{3}-9+6\sqrt{3}+9}{12-9}\)

\(=\frac{12\sqrt{3}}{3}=4\sqrt{3}\)

Answer 2

a. \(\left(\frac{\sqrt{x}+2}{x-9}-\frac{\sqrt{x}-2}{x+6\sqrt{x}+9}\right)\left(\sqrt{x}-\frac{9}{\sqrt{x}}\right)\)

= \(\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}-\frac{(\sqrt{x}-2)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}\right]\left(\frac{x}{\sqrt{x}}-\frac{9}{\sqrt{x}}\right)\)

= \(\left[\frac{x+3\sqrt{x}+2\sqrt{x}+6-x+3\sqrt{x}+2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}\right].\frac{x-9}{\sqrt{x}}\)

= \(\frac{10\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}.\frac{x-9}{\sqrt{x}}\)

= \(\frac{10}{\sqrt{x}+3}\)

b. \(\frac{3}{2\sqrt{3}+3}+\frac{3}{2\sqrt{3}-3}\)

= \(\frac{3\left(2\sqrt{3}-3\right)}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}+\frac{3\left(2\sqrt{3}+3\right)}{(2\sqrt{3}+3)\left(2\sqrt{3}-3\right)}\)

= \(\frac{6\sqrt{3}-9}{12-9}+\frac{6\sqrt{3}+9}{12-9}\)

= \(\frac{6\sqrt{3}-9+6\sqrt{3}+9}{3}\)

= \(\frac{12\sqrt{3}}{3}\)

= \(4\sqrt{3}\)