Question

[Kim Bắp]

giải giùm mình bài này:

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

a) tìm điều kiện

b) rút gọn

c) tính A biết x=\(6+4\sqrt{2}\)

2. Cho A=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

a) rút gọn

b) tìm x để \(A=\frac{-2}{5}\)

Answer 1

[Trang Hà]

a) ĐKXĐ \(\sqrt{x}\ne2\Leftrightarrow x\ne4\)

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

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

\(A=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

c) x = 6 + \(4\sqrt{2}\) = \(\left(2+\sqrt{2}\right)^2\)

=> A = \(\frac{\sqrt{\left(2+\sqrt{2}\right)^2}-4}{\sqrt{\left(2+\sqrt{2}\right)^2}-2}=\frac{\sqrt{2}-2}{\sqrt{2}}\)

Answer 2

[Trang Hà]

2.

a) đkxđ: \(x\ne4;x\ne9\)

A=\(\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

=\(\left(\frac{x-\sqrt{x}-6+x\sqrt{x}-9\sqrt{x}-2x+18-x\sqrt{x}+2x-4\sqrt{x}-8}{\left(x-4\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

\(\frac{x-14\sqrt{x}-5}{\left(x-4\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

b) A = -2/5

(k biết là do đề sai hay mình sai chứ đến đây nản quá! bạn làm nốt nhé!)