Question

6. Simplify: 3✓2/✓6-✓3 + 2✓3/✓6+2 - 4✓3/✓6-✓2​

Answer 1

Answer:

3√2/√6=3√2/√2*√3=√3

2√3/√6=√2

4√3/√6=2√2

therefore

√3-√3+√2+√2-2√2-√2

=-√2

Step-by-step explanation:

Answer 2

The answer is 3✓2/✓6-✓3 + 2✓3/✓6+2 - 4✓3/✓6-✓2​ = 0

Given:

3✓2/✓6-✓3 + 2✓3/✓6+2 - 4✓3/✓6-✓2​

To find:

Simplify: 3✓2/✓6-✓3 + 2✓3/✓6+2 - 4✓3/✓6-✓2​

Solution:

Given 3✓2/✓6-✓3 + 2✓3/✓6+2 - 4✓3/✓6-✓2​

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

Rationalize the denominators in each fraction as given below

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

⇒  $\frac{2\sqrt{3} }{\sqrt{6}+2 } [\frac{ \sqrt{6}-2}{ \sqrt{6}-2} ]$ =  $\frac{2\sqrt{18} - 4\sqrt{3} }{(\sqrt{6})^{2} +2^{2} }$ =  $\frac{2\sqrt{18} - 4\sqrt{3} }{2 }$ =  $\frac{2(\sqrt{18} - 2\sqrt{3}) }{2 }$ =  $\sqrt{18} - 2\sqrt{3}$

⇒  $\frac{4\sqrt{3} }{\sqrt{6}- \sqrt{2} } [ \frac{\sqrt{6}+ \sqrt{2} }{\sqrt{6}+\sqrt{2} } ]$ =  $\frac{4\sqrt{18} + 4\sqrt{6} }{(\sqrt{6})^{2} - (\sqrt{2} )^{2} }$ =  $\frac{4\sqrt{18} + 4\sqrt{6} }{4}$ =  $\frac{4(\sqrt{18} + \sqrt{6} )}{4}$ =  $\sqrt{18} + \sqrt{6}$  

From above calculations

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

= √12 - 2√3

= √4×3 - 2√3  

= 2√3 - 2√3  

= 0  

Therefore,

3✓2/✓6-✓3 + 2✓3/✓6+2 - 4✓3/✓6-✓2​ = 0

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