10.
Ta có: \(5=\sqrt{25}\)
\(2\sqrt{6}=\sqrt{2^2.6}=\sqrt{24}\)
Mà \(\sqrt{25}>\sqrt{24}\Rightarrow5>2\sqrt{6}\)
11.
\(A=2\sqrt{5}+3.\sqrt{3^2.5}=2\sqrt{5}+3.3.\sqrt{5}=2\sqrt{5}+9\sqrt{5}=11\sqrt{5}\)
12.
\(B=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a+3\sqrt{a}-\left(3\sqrt{a}-9\right)-\left(a-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{11}{a-9}\)
b.
\(B\) nguyên \(\Rightarrow a-9\inƯ\left(11\right)\)
\(\Rightarrow a-9\in\left\{-11;-1;1;11\right\}\)
\(\Rightarrow a\in\left\{-2;8;10;20\right\}\)
