Question

Rut gon:

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

Answer 1

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

Answer 2

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