Question

Cho biểu thức $A = \dfrac{\sqrt x + 2}{\sqrt x + 3} - \dfrac5{x + \sqrt x - 6} - \dfrac1{\sqrt x-2}$ với $x\ge 0$ và $x \ne 4$.

1. Rút gọn biểu thức $A$.

2. Tính giá trị của $A$ khi $x = 6+4\sqrt2$.

Answer 1

a, Với \(x\ge0,x\ne4\)

\(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

b, Ta có  \(x=6+4\sqrt{2}=2^2+4\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|=2+\sqrt{2}\)do \(2+\sqrt{2}>0\)

\(\Rightarrow A=\frac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\frac{-2+\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}+2}{2}=\frac{-2\left(\sqrt{2}-1\right)}{2}=1-\sqrt{2}\)

Answer 2

1, A = \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

2 , A = \(1-\sqrt{2}\)

Answer 3

Answer 4

1 .A=\(\dfrac{\left(\sqrt{x}+2\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-2\right)}-\dfrac{5}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)

  =\(\dfrac{x-2\sqrt{x}+2\sqrt{x}-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-2\right)}\)

    =\(\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-2\right)}\)

    =\(\dfrac{\left(\sqrt{x}-4\right).\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-2\right)}\)  =\(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

2. Thay x=\(6+4\sqrt{2}\) vào bt A ta đc:

\(\Rightarrow\dfrac{\sqrt{6+4\sqrt{2}}-4}{\sqrt{6+4\sqrt{2}}-2}\)