Question

Giúp mik với

Answer 1

\(\left(\dfrac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+\sqrt{a}^3}{1+\sqrt{a}}-\sqrt{a}\right)=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)=\left(1+2\sqrt{a}+a\right)\left(1-2\sqrt{a}+a\right)=\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2=\left(a-1\right)^2\)

Answer 2

Ta có: \(D=\left(\dfrac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\)

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

\(=a^2-2a+1\)