Question

Cho \(A=\frac{x+3\sqrt{x}}{x-\sqrt{x}+1}\)và \(B=\frac{6\sqrt{x}}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}-2}{\sqrt{x}+3}\) \(\left(x\ge9;x\ne9\right)\)

1) Tính giá trị của A khi x=\(\frac{1}{9}\)

2) Rút gọn biểu thức B.

3) Đặt P=AB. Hãy so sánh P với 2

Answer 1

1) Thay \(x=\frac{1}{9}\) vào biểu thức \(A=\frac{x+3\sqrt{x}}{x-\sqrt{x}+1}\), ta được:

\(A=\left(\frac{1}{9}+3\cdot\sqrt{\frac{1}{9}}\right):\left(\frac{1}{9}-\sqrt{\frac{1}{9}}+1\right)\)

\(=\left(\frac{1}{9}+3\cdot\frac{1}{3}\right):\left(\frac{1}{9}-\frac{1}{3}+1\right)\)

\(=\left(\frac{1}{9}+1\right):\frac{7}{9}\)

\(=\frac{10}{9}:\frac{7}{9}=\frac{10}{9}\cdot\frac{9}{7}=\frac{10}{7}\)

Vậy: Khi \(x=\frac{1}{9}\) thì \(A=\frac{10}{7}\)

2) Ta có: \(B=\frac{6\sqrt{x}}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

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

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

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

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

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