Question

A= \(\left(1-\frac{4\sqrt{x}}{x-1}+\frac{1}{\sqrt{x}-1}\right):\frac{x-2\sqrt{x}}{x-1}\)\(\left(x>0;x\ne1;x\ne4\right)\)

a, Rút gọn

b, Tìm x để A= 1/2

Answer 1

a)

\(A=\left(\frac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{x-1}{x-2\sqrt{x}}\\ =\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\right)\cdot\frac{x-1}{x-2\sqrt{x}}\\ =\frac{x-3\sqrt{x}}{x-1}\cdot\frac{x-1}{x-2\sqrt{x}}\\ =\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{\sqrt{x}-3}{\sqrt{x}-2}\)

b)

\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{1}{2}\\ \Leftrightarrow2\left(\sqrt{x}-3\right)=\sqrt{x}-2\\ \Leftrightarrow2\sqrt{x}-6=\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}=4\\ \Leftrightarrow x=16\left(tm\right)\)