Cho biểu thức P = (1/ a-1+ 3 căn bậc hai a + 5/ a căn bậc hai a - a - căn bậc hai a + 1) [(căn bậc hai a +1)^2/4 căn bậc hai a -1]

a) Với $a>0,a\ne 1$ ta có:

$$\begin{gathered} P=\left[\frac{1}{a-1}+\frac{3\sqrt{a}+5}{a\left(\sqrt{a}-1\right)-\left(\sqrt{a}-1\right)}\right].\left[\frac{(a+2\sqrt{a}+1)-4\sqrt{a}}{4\sqrt{a}}\right]=\left[\frac{1}{a-1}+\frac{3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right].\frac{a-2\sqrt{a}+1}{4\sqrt{a}} \\ =\frac{\sqrt{a}-1+3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\frac{{\left(\sqrt{a}-1\right)}^2}{4\sqrt{a}}=\frac{4\sqrt{a}+4}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\frac{{\left(\sqrt{a}-1\right)}^2}{4\sqrt{a}}=\frac{4\sqrt{a}+4}{a-1}.\frac{\sqrt{a}-1}{4\sqrt{a}} \\ =\frac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}-1}{4\sqrt{a}}=\frac{1}{\sqrt{a}} \end{gathered}$$