Question

Cho biểu thức

P= \(\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}\right)\times\dfrac{a+2\sqrt{a}+1}{\sqrt{a}-1}\) với \(a>0\) và \(a\ne1\)

a) Rút gọn biểu thức P

b) Tính giá trị của P khi \(a=2\sqrt{3-\sqrt{5}}\times\left(3+\sqrt{5}\right)\times\left(\sqrt{10}-\sqrt{2}\right)\)

Answer 1

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

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

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

b/

\(a=2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(a=\sqrt{3-\sqrt{5}}\left(6+2\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\)

\(a=\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)\)

\(a=\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2\)

\(a=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)

\(\Rightarrow P=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{16}+1}{\sqrt{16}}=\dfrac{4+1}{4}=\dfrac{5}{4}\)