Question

cho biểu thức

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

với \(x>0,x\ne1\)

a. rút gọn M

b. tìm giá trị của x sao cho M>1

Answer 1

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

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

\(M=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)}{\sqrt{x}}=\frac{4}{\sqrt{x}+1}\)

\(M>1\Rightarrow\frac{4}{\sqrt{x}+1}>1\Rightarrow4>\sqrt{x}+1\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)

Vậy để M>1 thì \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)