Question

Rut gon bieu thuc sau:

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

Answer 1

điều kiện : \(x>0;x\ne1\)

ta có : \(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)

\(\Leftrightarrow A=\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)

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

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