Question

cho P= \(\left[{}\begin{matrix}1\\\sqrt{X}-1\end{matrix}\right.-\dfrac{1}{\sqrt{X}}]:\left[{}\begin{matrix}\sqrt{X}+1\\\sqrt{X}-2\end{matrix}\right.-\dfrac{\sqrt{X}+2}{\sqrt{X}-1}]\)

a) tim DKXD

b.rút gọn P

c.tìm x để P=\(\dfrac{1}{4}\)

Answer 1

a/ ĐKXĐ: \(x>0,x\ne1,x\ne2\)

b/

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

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

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

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

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

c/ Với \(x>0,x\ne1,x\ne2\)

Để P=\(\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=8\)

\(\Leftrightarrow x=64\left(tm\right)\)

Vậy để \(P=\dfrac{1}{4}\) thì x=64