Question

Cho biểu thức P=\(\left(\dfrac{1}{a}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{a+\sqrt{a}}\)

a. Rút gọn P

Answer 1

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

Answer 2

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