Question

Rút gọn P=(1+\(\dfrac{a+\sqrt{a}}{1+\sqrt{a}}\)) (1-\(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\))

giúp mình vs ạ

Answer 1

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

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

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

P = \(\dfrac{a+2\sqrt{a}+1}{1+\sqrt{a}}.\dfrac{-a+2\sqrt{a}-1}{\sqrt{a}-1}\)

P = \(\dfrac{a+2\sqrt{a}+1}{1+\sqrt{a}}.\dfrac{a-2\sqrt{a}+1}{1-\sqrt{a}}\)

P = \(\dfrac{\left(1+\sqrt{a}\right)^2}{1+\sqrt{a}}.\dfrac{\left(1-\sqrt{a}\right)^2}{1-\sqrt{a}}\) = \(\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

Answer 2

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

\(=\left[1+\dfrac{\sqrt{a}\left(1+\sqrt{a}\right)}{1+\sqrt{a}}\right]\left[1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a-1}}\right]\)

= \(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)