Question

[Đặng Phương Thảo]

1.Cho C=\(\frac{2\sqrt{x}-9}{x+5\sqrt{x}+6}\)\(-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)\(-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

a, Rút gon C

b,Tìm x nguyên để b nguyên

GIÚP MÌNH VỚIIIII

Answer 1

[Hoàng Thị Lan Hương]

a. \(C=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

b. C=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

C nguyên \(\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\Rightarrow\sqrt{x}-3\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)

Vậy \(x\in\left\{1;4;16;25;49\right\}\)thì C nguyên