Question

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

a. Rút gọn A

b. Tìm x để A<0

c. Tìm x nguyên để A có giá trị nguyên

Answer 1

ĐKXĐ:...

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

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

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

Để \(A< 0\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Rightarrow x< 1\)

c/ \(A=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)

Để A nguyên \(\Rightarrow\sqrt{x}-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(\sqrt{x}-1=-2\Rightarrow\sqrt{x}=-1< 0\left(l\right)\)

\(\sqrt{x}-1=-1\Rightarrow x=0\)

\(\sqrt{x}-1=1\Rightarrow x=4\)

\(\sqrt{x}-1=2\Rightarrow x=9\)