If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, then find the value of
$\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}$
We have, $\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}$
$=\frac{4(3 \sqrt{3}+2 \sqrt{2})+3(3 \sqrt{3}-2 \sqrt{2})}{(3 \sqrt{3}-2 \sqrt{2})(3 \sqrt{3}+2 \sqrt{2})}$
$=\frac{12 \sqrt{3}+8 \sqrt{2}+9 \sqrt{3}-6 \sqrt{2}}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}}$ [using identity, $(a+b)(a-b)=a^{2}-b^{2}$ ]
$=\frac{21 \sqrt{3}+2 \sqrt{2}}{27-8}=\frac{21 \sqrt{3}+2 \sqrt{2}}{19}$
$=\frac{21 \times 1.732+2 \times 1.414}{19}$ [put $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$ ]
$=\frac{36.372+2.828}{19}=\frac{39.2}{19}=2.06316 \approx 2.063$
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