Question

6/(2sqrt(3) - sqrt(6)) + 6/(sqrt(3) + sqrt(2)) + (4sqrt(3))/(sqrt(6) - sqrt(2))​

Answer 1

First Second

This is your answer. Hope it will help you a lot

Answer 2

Answer:

The result is 8√3+2√6-3√2

Step-by-step explanation:

Given that 6/(2√3-√6) + 6/(√3+√2) + 4√3/√6-√2

We have to simplify the above expression

Take individual expressions and rationalization .

in this way the denominator only contains a rational number.

suppose a/√a-√b = $\frac{a}{\sqrt{a}-\sqrt{b} } *\frac{\sqrt{a}+\sqrt{b} }{\sqrt{a} +\sqrt{b} }$

(√a+√b)(√a-√b) = (√a)²-(√b)² = a-b

$\frac{6}{2\sqrt{3} -6}$ $=\frac{6}{2\sqrt{3}-\sqrt{6} } *\frac{2\sqrt{3}+\sqrt{6} }{2\sqrt{3} +\sqrt{6} } = \frac{6(2\sqrt{3}+\sqrt{6}) }{(2\sqrt{3})^{2}-(\sqrt{6} )^{2} } =\frac{6(2\sqrt{3}+\sqrt{6}) }{12-6} =\frac{6(2\sqrt{3}+\sqrt{6}) }{6} =2\sqrt{3} +\sqrt{6}$

$\frac{6}{\sqrt{3}+\sqrt{2} }$$=\frac{6}{\sqrt{3}+\sqrt{2} } *\frac{\sqrt{3}-\sqrt{2} }{\sqrt{3 }-\sqrt{2} } =\frac{6(\sqrt{3} -\sqrt{2)} }{(\sqrt{3})^{2} -(\sqrt{2})^{2} } =\frac{6(\sqrt{3}-\sqrt{2}) }{3-2} =6(\sqrt{3}-\sqrt{2} )$

$\frac{4\sqrt{3} }{\sqrt{6}-\sqrt{2} }$$=\frac{4\sqrt{3} }{\sqrt{6} -\sqrt{2} } *\frac{\sqrt{6}+\sqrt{2} }{\sqrt{6}+\sqrt{2} } =\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2}) }{(\sqrt{6} )^{2}-(\sqrt{2})^{2} } =\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2}) }{6-2} =\frac{4\sqrt{3} (\sqrt{6}+\sqrt{2}) }{4} =\sqrt{3} (\sqrt{6} +\sqrt{2} )$

Now the expression is

$\frac{6}{2\sqrt{3} -6}$ + $\frac{6}{\sqrt{3}+\sqrt{2} }$ +$\frac{4\sqrt{3} }{\sqrt{6}-\sqrt{2} }$ = $2\sqrt{3} +\sqrt{6} +6(\sqrt{3} -\sqrt{2} )+\sqrt{3} (\sqrt{6} +\sqrt{2} )$

                                    $=2\sqrt{3} +\sqrt{6} +6\sqrt{3} -6\sqrt{2} +(\sqrt{3} *\sqrt{6} )+(\sqrt{3} *\sqrt{2} )$

                                    =$2\sqrt{3} +\sqrt{6} +6\sqrt{3} -6\sqrt{2} +\sqrt{18} +\sqrt{6}$

                                    $=2\sqrt{3} +2\sqrt{6} +6\sqrt{3} -6\sqrt{2} +3\sqrt{2}$

                                   $=8\sqrt{3} +2\sqrt{6} -3\sqrt{2}$

Therefore, the result is 8√3+2√6-3√2.