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 gọn Q

b So sánh Q với 1

Answer 1

a) ĐK: \(x>0;x\ne1\)

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

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

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

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

b) Ta có: \(Q=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{a}}{\sqrt{a}}+\dfrac{1}{\sqrt{a}}=1+\dfrac{1}{\sqrt{a}}>1\)

Answer 2

Sửa lại:

a)Đk:x>0