Question

Bài 2: Cho A= \(\frac{2\sqrt{x}}{\sqrt{x}+3}\)và B= \(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1}{x+\sqrt{x}}\)
a) Rút gọn B
b) Tìm x để \(A>\frac{2}{3}\)
c) Tìm x để \(\frac{A}{B}\)nhận giá trị nguyên

Answer 1

a, Với x > 0 

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

b, Ta có : \(A>\frac{2}{3}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{2}{3}>0\Leftrightarrow\frac{3\sqrt{x}-2\sqrt{x}-2}{3\left(\sqrt{x}+1\right)}>0\)

\(\Rightarrow\sqrt{x}-2>0\Leftrightarrow x>4\)

c, \(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}+3}{2\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}+2}=\frac{2\sqrt{x}+6}{2\sqrt{x}+2}=1+\frac{4}{2\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+1}\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\)

\(\sqrt{x}+1\)	1	2
\(\sqrt{x}\)	0 (loại )	1
x	loại	1