Question

[Hảo Hảo]

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

a) Rút gọn M

b) tìm x để M<0

c) tìm các giá trị nguyên của x để M nguyên dương

Answer 1

[tran nguyen bao quan]

ĐKXĐ:\(x\ge0,x\ne1\)

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

b) Ta có \(M< 0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)

Mà \(\sqrt{x}+1>0\)

Suy ra \(\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

Kết hợp với ĐKXĐ

Vậy \(0\le x< 1\) thì M<0

c) Ta có \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Vậy để M nguyên dương thì \(\left\{{}\begin{matrix}\sqrt{x}-1\inƯ\left(2\right)\\M>0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x}-1\in\left\{\pm1;\pm2\right\}\\M>0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=0\\\sqrt{x}=-1\left(l\right)\\\sqrt{x}=3\end{matrix}\right.\\M>0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=4\left(tm\right)\\x=1\left(ktm\right)\\x=0\left(ktm\right)\end{matrix}\right.\\M>0\end{matrix}\right.\)

Vậy x=4 thì M nguyên dương