Question

P=\(\left(\frac{1}{\sqrt{a}+2}+\frac{1}{\sqrt{a}-2}\right).\frac{\sqrt{a}-2}{\sqrt{a}}\)
a,Rút gọn
b,Tìm a để P>\(\frac{1}{3}\)

Answer 1

ĐKXĐ: \(a>0;a\ne4\)

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

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

\(P>\frac{1}{3}\Leftrightarrow\frac{2}{\sqrt{a}+2}>\frac{1}{3}\Leftrightarrow6>\sqrt{a}+2\)

\(\Rightarrow\sqrt{a}< 4\Rightarrow a< 16\)

Vậy \(\left\{{}\begin{matrix}0< a< 16\\a\ne4\end{matrix}\right.\)