Question

Cho biểu thức: A = \(\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\) với x > 0 và x \(\ne\) 1

a) Rút gọn A

b) Tìm x nguyên để A nhận giá trị nguyên

Answer 1

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

b) Ta có \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)

Vậy để A nhận giá trị nguyên thì \(\sqrt{x}-1\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)

Vì \(\sqrt{x}-1\ge-1\)

Suy ra \(\left[{}\begin{matrix}\sqrt{x}-1=1\\\sqrt{x}-1=-1\\\sqrt{x}-1=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(ktm\right)\\x=9\left(tm\right)\end{matrix}\right.\)

Vậy x=4 hoặc x=9 thì A nhận giá trị nguyên

Answer 2

bài 1:a)

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

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

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

A=\(\frac{2\left(x-1\right)}{2\left(x-1\right)^2}\) <=> A=\(\frac{1}{x-1}\)