Question

Compare the following two numbers :$a=\frac{9}{\sqrt{11}-\sqrt{2}}$ and $b=\frac{6}{3\sqrt{3}}$

Answer 1

a=$\frac{9}{\sqrt{11}-\sqrt{2}}$

$\frac{9}{\sqrt{11}-\sqrt{2}}\times \frac{\sqrt{11}+\sqrt{2}}{\sqrt{11}+\sqrt{2}}$

$\frac{9(\sqrt{11}+\sqrt{2})}{11-2}$ use $a^2-b^2=(a+b)(a-b)$

$\frac{9(\sqrt{11}+\sqrt{2})}{9}=(\sqrt{11}+\sqrt{2})$

$a=(\sqrt{11}+\sqrt{2})$

b=$\frac{6}{\sqrt{3}}$

$\frac{6}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$b=\frac{6\sqrt{3}}{9}=\frac{2\sqrt{3}}{3}$

so $a>b$ . answer.