Question

simplify under root 6 upon under root 2 + under root 3 + 3 root 2 upon under root 6 + under root 3 minus 4 root 3 upon under root 6 + under root 2

Answer 1

Answer: The answer is 0.

Step-by-step explanation:

Given expression is,

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

By rationalizing each term,

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

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

$=-\sqrt{12} + \sqrt{18} + \sqrt{12} - \sqrt{6} - \sqrt{18} + \sqrt{6}$

$=0$

Hence,

$\frac{\sqrt{6} }{\sqrt{2} + \sqrt{3} } + \frac{3\sqrt{2} }{\sqrt{6}+\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6}+\sqrt{2} }=0$

Answer 2

Answer:

$\frac{\sqrt{6} }{\sqrt{2} + \sqrt{3} } + \frac{3\sqrt{2} }{\sqrt{6}+\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6}+\sqrt{2}}=0$

Step-by-step explanation:

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

To find : Simplify the expression ?

Solution :

Step 1 - Write the expression,

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

Step 2 - Rationalize each term,

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

Step 3 - Simplify,

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

$=-\sqrt{12} + \sqrt{18} + \sqrt{12} - \sqrt{6} - \sqrt{18} + \sqrt{6}$

$=0$

Therefore, $\frac{\sqrt{6} }{\sqrt{2} + \sqrt{3} } + \frac{3\sqrt{2} }{\sqrt{6}+\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6}+\sqrt{2}}=0$