Math, asked by faizahnazreen, 1 year ago

Simplify the following

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Answers

Answered by prashilpa
2

Answer:

(2√6 + 3√2 -6) / 2

Step-by-step explanation:

(3√2)/( √6 - √3)  + (2 √3)/( √6 + √2)  - (6 √2)/( √6 - √2)

Each of the denominator is in the form of A + B or A – B form.  

If the Denominator is in the form of A – B, multiply the numerator and denominator with A + B.  

If the Denominator is in the form of A + B, multiply the numerator and denominator with A – B.  

(A + B)(A – B) = A^2 – B^2

Thus gives.

(3√2) ( √6 + √3)/( 6 - 3)  + (2 √3) ( √6 - √2)/( 6 - 2)  - (6 √2) ( √6 + √2)/( 6 - 2)

= (3√2√6 + 3√2√3)/(2)  + (2√3√6 - 2√3√2)/( 4)  - (6√2√6 + 6√2√2)/( 4)

Multiplying first fraction numarator and denominator by 2

= 2(6√3 + 3√6)/4 + (6√2 - 2√6)/4 – (12√3 + 12)/4

= (12√3 + 6√6 + 6√2 - 2√6 –12√3 -12)/4

= (4√6 + 6√2 -12)/4

= (2√6 + 3√2 -6)/2

Answered by hukam0685
2
To simplify:

 \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{2 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } - \frac{6 \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \\ \\ = \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{2 \sqrt{3} ( \sqrt{6} - \sqrt{2} ) - 6 \sqrt{2} ( \sqrt{6} + \sqrt{2}) }{( \sqrt{6} + \sqrt{2})( \sqrt{6} - \sqrt{2} )} \\ \\ \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{2 \sqrt{18} - 2 \sqrt{6} - 6 \sqrt{12} - 12 }{( { \sqrt{6} )}^{2} - ( { \sqrt{2} )}^{2} } \\ \\ = \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{6 \sqrt{2} - 2 \sqrt{6} - 12 \sqrt{3} - 12}{6 - 2} \\ \\ = \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{6 \sqrt{2} - 2 \sqrt{6} - 12 \sqrt{3} - 12}{4} \\ \\ = \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } + \frac{3\sqrt{2} - \sqrt{6} - 6 \sqrt{3} - 6}{2} \\ \\ \frac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \frac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } + \frac{3\sqrt{2} - \sqrt{6} - 6 \sqrt{3} - 6}{2} \\ \\ = \frac{3 \sqrt{2}( \sqrt{6} + \sqrt{3} )}{( { \sqrt{6} )}^{2} - ( { \sqrt{3} )}^{2} } + \frac{3\sqrt{2} - \sqrt{6} - 6 \sqrt{3} - 6}{2} \\ \\ = \frac{3 \sqrt{12} + 3 \sqrt{6} }{6 - 3} + \frac{3\sqrt{2} - \sqrt{6} - 6 \sqrt{3} - 6}{2} \\ \\ \frac{6 \sqrt{3} + 3 \sqrt{6} }{3} + \frac{3\sqrt{2} - \sqrt{6} - 6 \sqrt{3} - 6}{2} \\ \\ = \frac{12 \sqrt{3} + 6 \sqrt{6} + 9 \sqrt{2} - 3 \sqrt{6} - 18 \sqrt{3} - 18}{6} \\ \\ = \frac{3 \sqrt{6} - 6 \sqrt{3} + 9 \sqrt{2} - 18}{6} \\ \\ = \frac{ \sqrt{6} - 2 \sqrt{3} + 3\sqrt{2} - 6}{2} \\ \\

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

amitnrw: from third last step there is a mistake you cancelled 3 from numerator and denominator but you mentioned 6/3 = 3 instead of 2 in denominator
hukam0685: yes,by mistaken
faizahnazreen: Tysmmmmm
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