Question

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

a) rút gọn P

Answer 1

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

\(=\frac{a+\sqrt{a}+1}{a+1}:\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)

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

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

Answer 2

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

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

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

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