Question

A=\(\left(\frac{1}{\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+1}\right)\); B=\(\frac{\sqrt{a}}{a+\sqrt{a}}\) (a>0)
a,Rút gọn bt P=A:B
b,Tính gt của P khi a=\(\frac{2}{\sqrt{5}-1}-\frac{2}{\sqrt{5}+1}\)
c,Tính gt nhỏ nhất của P

Answer 1

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

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

\(a=\frac{2}{\sqrt{5}-1}-\frac{2}{\sqrt{5}+1}=\frac{2\sqrt{5}+2-2\sqrt{5}+2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{4}{4}=1\)

\(\Rightarrow P=3\)

\(P=\frac{a+\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}+1\ge2\sqrt{\frac{\sqrt{a}}{\sqrt{a}}}+1=3\)

\(\Rightarrow P_{min}=3\) khi \(a=1\)