Question

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

a) rút gọn M

b) tìm x để M = -1/2

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

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

\(=\dfrac{x}{\sqrt{x}-1}\)

b.

\(M=-\dfrac{1}{2}\Leftrightarrow\dfrac{x}{\sqrt{x}-1}=-\dfrac{1}{2}\)

\(\Rightarrow2x=1-\sqrt{x}\Rightarrow2x+\sqrt{x}-1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vn\right)\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{4}\) (tm)

Answer 2

ĐKXĐ : \(x>0;x\ne1\)

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

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

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

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

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

\(=\dfrac{x}{\sqrt{x}-1}\)

Vậy : \(M=\dfrac{x}{\sqrt{x}-1}\)

 

b. Để \(M=-\dfrac{1}{2}\) thì : \(\dfrac{x}{\sqrt{x}-1}=-\dfrac{1}{2}\)

\(\Leftrightarrow-2x=\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}+2x-1=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(tmđk\right)\\x\in\varnothing\end{matrix}\right.\)

Vậy : \(M=-\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)