Question

Cho P = (\(\dfrac{1}{\sqrt{x}-1 }\) - \(\dfrac{1}{\sqrt{x}}\))(\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\))

a. Tìm đkxđ và rút gọn P

b. Tìm x để P = \(\dfrac{1}{4}\) 

Answer 1

Điều kiện: x>2

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

P= \(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b) P= \(\dfrac{1}{4}\)

⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)

⇔\(4\sqrt{x}-8=3\sqrt{x}\)

⇔\(\sqrt{x}=8\)

⇔x=64 (TM) 

Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)