Question

Simplify 3 root 2 by root 6 minus root 3 minus 4 root 3 by root 6 + root 3 + 2 root 3 by 6 root + 2

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\sqrt{12} +\sqrt{6} - 4\sqrt{2} + 4 +{\sqrt{18} - 2\sqrt{3}$ is the required value of

$\frac{3\sqrt{2} }{\sqrt{6}-\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6+ \sqrt{3} } } +\frac{2\sqrt{3} }{\sqrt{6} + 2 }$.

Step-by-step explanation:

Explanation:

Given in the question that, $\frac{3\sqrt{2} }{\sqrt{6}-\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6+ \sqrt{3} } } +\frac{2\sqrt{3} }{\sqrt{6} + 2 }$

First we rationalize the given numerical one by one.

Step 1:

First we rationalize $\frac{3\sqrt{2} }{\sqrt{6}-\sqrt{3} }$,

So, multiplying numerator and denominator by $\sqrt{6} +\sqrt{3}$,

⇒$\frac{3\sqrt{2} }{\sqrt{6}-\sqrt{3} }$ × $\frac{\sqrt{6} +\sqrt{3}}{\sqrt{6} +\sqrt{3}}$

⇒$\frac{3\sqrt{2}(\sqrt{6} +\sqrt{3}) }{(\sqrt{6} -\sqrt{3})(\sqrt{6} +\sqrt{3})}$ = $\frac{3(\sqrt{12} +\sqrt{6} )}{6 - 3}$ = $\frac{3(\sqrt{12} +\sqrt{6} )}{3}$  [where ($a^2 - b^2 = (a + b)(a - b)$)]

⇒ $(\sqrt{12} +\sqrt{6} )$ .........(i)

Now we rationalize $\frac{4\sqrt{3} }{\sqrt{6+ \sqrt{3} } }$,

Multiply numerator and denominator by $\sqrt{6} - \sqrt{3}$

⇒$\frac{4\sqrt{3} }{\sqrt{6+ \sqrt{3} } }$ ×$\frac{\sqrt{6} - \sqrt{3}}{\sqrt{6} -\sqrt{3}}$

⇒$\frac{4\sqrt{3}(\sqrt{6} - \sqrt{3} ) }{6 - 3}$ = $\frac{4\sqrt{3}\sqrt{3} (\sqrt{2} - 1 ) }{3}$ = $(4\sqrt{2} - 4 )$. ........(ii)

Similarly, on rationalizing $\frac{2\sqrt{3} }{\sqrt{6} + 2 }$ we get,

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

⇒$\frac{2\sqrt{3}(\sqrt{6} - 2) }{2}$ = ${(\sqrt{18} - 2\sqrt{3} )$.........(iii)

Step 2:

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

On putting the values  we get, $(\sqrt{12} +\sqrt{6} ) - (4\sqrt{2} - 4 ) +{(\sqrt{18} - 2\sqrt{3} )$

⇒$\sqrt{12} +\sqrt{6} - 4\sqrt{2} + 4 +{\sqrt{18} - 2\sqrt{3}$

Final answer:

Hence, the value of $\frac{3\sqrt{2} }{\sqrt{6}-\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6+ \sqrt{3} } } +\frac{2\sqrt{3} }{\sqrt{6} + 2 }$ is

$\sqrt{12} +\sqrt{6} - 4\sqrt{2} + 4 +{\sqrt{18} - 2\sqrt{3}$.

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