Question : Given that $\sqrt3=1.732$, the value of $\frac{3+\sqrt6}{5\sqrt3-2\sqrt{12}-\sqrt{32}+\sqrt{50}}$ is:
Option 1: 4.899
Option 2: 2.551
Option 3: 1.414
Option 4: 1.732
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Correct Answer: 1.732
Solution :
Given:
$\sqrt{3}=1.732$
$\frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}$
Now evaluate:
$= \frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{4\times3}-\sqrt{16\times2}+\sqrt{25\times2}}$
$=\frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}$
$=\frac{3+\sqrt{6}}{5(\sqrt{3}+\sqrt{2})-4(\sqrt{3}+\sqrt{2})}$
$=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}$
Now multiply and divide with $\sqrt{3}-\sqrt{2}.$
$=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{3\sqrt{3}-3\sqrt{2}+\sqrt{18}-\sqrt{12}}{3-2}$
$=\frac{3\sqrt{3}-3\sqrt{2}+\sqrt{9\times2}-\sqrt{4\times3}}{1}$
$=3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}$
$=3\sqrt{3}-2\sqrt{3}$
$=\sqrt{3}(3-2)$
$=\sqrt{3}$
$=1.732$
Hence, the correct answer is 1.732.
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