Question

Rút gọn P:

\(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}-\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)

Answer 1

ĐK: \(ab\ge0\)

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

\(P=-1.\)

Answer 2

\(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}-\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)\(P=\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}+\dfrac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}-\dfrac{ab-1}{ab-1}\right]:\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}-\dfrac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}+\dfrac{ab-1}{ab-1}\right]\)\(P=\dfrac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)+\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)-\left(ab-1\right)}{ab-1}:\dfrac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)-\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)+\left(ab-1\right)}{ab-1}\)\(P=\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{ab-1}:\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{ }\)\(P=\dfrac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}:\dfrac{-2\sqrt{a}-2}{ab-1}\)
\(P=\dfrac{2\sqrt{ab}\left(\sqrt{a}+1\right)}{ab-1}.\dfrac{ab-1}{-2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)