Question

a) Rút gọn A \(\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{\sqrt{a}}{a-1}\right):\left(\frac{2}{a}-\frac{2-a}{a\sqrt{a}+a}\right)\)

b) tìm a để A = \(\frac{6}{5}\)

Answer 1

mình nhấn nhầm đoạn cuối \(\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{\sqrt{a}}{a-1}\right)\left(\frac{a+\sqrt{a}}{2+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}}{a-1}\right)\left(\frac{a+\sqrt{a}}{2+\sqrt{a}}\right)=\frac{\sqrt{a}\left(\sqrt{a}+2\right)\left(a+\sqrt{a}\right)}{\left(a-1\right)\left(2+\sqrt{a}\right)}\) \(=\frac{\sqrt{a}\left(\sqrt{a}+a\right)}{a-1}=\frac{a+a\sqrt{a}}{a-1}=\frac{a\left(a+\sqrt{a}\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{a}{\sqrt{a}-1}\)

Answer 2

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