Question
Simplify:
$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}$
$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}$
✓
Answer
We know that rationalization factor for $3\sqrt2+2\sqrt3$ and $\sqrt3-\sqrt2$ are $3\sqrt2-2\sqrt3$ and $\sqrt3+\sqrt2$ respectively. We will multiply numerator and denominator of the given expression $\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}$ and $\frac{\sqrt{12}}{\sqrt3-\sqrt2}$ by $3\sqrt2-2\sqrt3,$ and $\sqrt3+\sqrt2$ respectively. to get
$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}\times\frac{3\sqrt2-2\sqrt3}{3\sqrt2-2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(3\sqrt2\big)^2+\big(2\sqrt3\big)^2-2\times3\sqrt2\times2\sqrt3}{\big(3\sqrt2\big)^2-\big(2\sqrt3\big)^2}+\frac{\sqrt{36}+\sqrt{24}}{\big(\sqrt3\big)^2-\big(\sqrt2\big)^2}$
$=\frac{18+12-12\sqrt6}{18-12}+\frac{6+\sqrt{24}}{3-2}$
$=\frac{30-12\sqrt6+36+12\sqrt6}{6}$
$=\frac{66}{6}$
$=11$
Hence the given expression is simplified to 11.
$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}\times\frac{3\sqrt2-2\sqrt3}{3\sqrt2-2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(3\sqrt2\big)^2+\big(2\sqrt3\big)^2-2\times3\sqrt2\times2\sqrt3}{\big(3\sqrt2\big)^2-\big(2\sqrt3\big)^2}+\frac{\sqrt{36}+\sqrt{24}}{\big(\sqrt3\big)^2-\big(\sqrt2\big)^2}$
$=\frac{18+12-12\sqrt6}{18-12}+\frac{6+\sqrt{24}}{3-2}$
$=\frac{30-12\sqrt6+36+12\sqrt6}{6}$
$=\frac{66}{6}$
$=11$
Hence the given expression is simplified to 11.
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