Question

chứnng minh rằng :

a, điều kiện : a> hoặc = 0, a#1\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)

Answer 1

Ta có:

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

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

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

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

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

\(=-\left(a-1\right)\)

\(=1-a\)

\(\rightarrowđpcm\)

Answer 2

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

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