Question

Cho biểu thức:

A=\(\left(\dfrac{x-3\sqrt[]{x}}{x-9}-1\right)\): \(\left(\dfrac{9-x}{\left(\sqrt[]{x}+3\right)\left(\sqrt[]{x}-2\right)}+\dfrac{\sqrt[]{x}-3}{\sqrt[]{x}-2}-\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}+3}\right)\)

a.Tìm x để A có nghĩa

b.Rút gọn A.Tính A khi x=3 - 2\(\sqrt[]{2}\)

c.Tìm x để A < 1

Answer 1

a: ĐKXĐ: x>=0; \(x\notin\left\{4;9\right\}\)

b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\dfrac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

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

\(=\dfrac{3}{\sqrt{x}+2}\)

Thay \(x=3-2\sqrt{2}\) vào A, ta được:

\(A=\dfrac{3}{\sqrt{2}-1+2}=\dfrac{3}{\sqrt{2}+1}=3\sqrt{2}-3\)

c: Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{3-\sqrt{x}-2}{\sqrt{x}+2}< 0\)

\(\Leftrightarrow1-\sqrt{x}< 0\)

hay 0<x<1