Question

Answer 1

\(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\) (ĐK: \(a\ge0;a\ne1\))

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

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

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

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

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

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

Answer 2

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

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

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