Question

1. Cho A = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)    
a) Rút gọn A
b) Tìm A khi x = \(\frac{1}{9}\)
c) Tìm x\(\in\)Z để A\(\in\)Z

 

 

Answer 1

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

a) \(A=\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(A=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

Answer 2

b) Thay \(x=\frac{1}{9}\)vào A, ta có:

\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=\frac{\frac{1}{3}+1}{\frac{1}{3}-3}=\frac{\frac{4}{3}}{\frac{-8}{3}}=-\frac{1}{2}\)

Answer 3

c) \(A\in Z\Rightarrow\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\Leftrightarrow\sqrt{x}-3\)là ước số của 4 gồm:\(1;2;4;-1;-2;-4\)

\(\sqrt{x}-3=1\Leftrightarrow x=16\)(chọn)

\(\sqrt{x}-3=2\Leftrightarrow x=25\)(chọn)

\(\sqrt{x}-3=4\Leftrightarrow x=49\)(chọn)

\(\sqrt{x}-3=-1\Leftrightarrow x=4\)(loại)

\(\sqrt{x}-3=-2\Leftrightarrow x=1\)(chọn)

\(\sqrt{x}-3=-4\Leftrightarrow\sqrt{x}=-1\)(vô lý)

Vậy với \(x=1;16;25;49\)thì \(A\in Z\)