Question

cho biểu thức C= \(\dfrac{\sqrt{x}+1}{\sqrt{x-1}}-\dfrac{4\sqrt{x}}{x-1}\)vs x\(\ge0\)  x\(\ne0\)

a) rút gọn c

b tìm x biết c= \(\dfrac{1}{3}\)

Answer 1

a) \(C=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b) \(C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{x}+1=3\sqrt{x}-3\Leftrightarrow2\sqrt{x}=4\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)