Question

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

a. Tìm điều kiện và rút gon Q

b. So sánh Q với 1

Answer 1

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)

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

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