Math, asked by Devchauhanuuu, 1 year ago

simplify the following

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Answered by Srsukh
1
here is your 1st 2nd and 3rd answer
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Devchauhanuuu: thanks Bhai
Answered by HimanshuR
1
i)
(4 \sqrt{3}  - 2 \sqrt{2} )(3 \sqrt{2}  + 4 \sqrt{3} ) \\  = 4 \sqrt{3}  \times 3 \sqrt{2} + 4 \sqrt{3}   \times 4 \sqrt{3}  \\  - 2 \sqrt{2}  \times 3 \sqrt{2}  - 2 \sqrt{2}  \times 4 \sqrt{3}  \\  = 12 \sqrt{6}  + 48 - 12 - 8 \sqrt{6}  \\  = 4 \sqrt{6}  + 36
ii)
(2 +  \sqrt{3} )( 3 +  \sqrt{5}) \\  = 2 \times3 + 2 \times  \sqrt{5}   +  \sqrt{3}  \times 3 +  \sqrt{3} \times  \sqrt{5} \\  = 6 + 2 \sqrt{5} + 3 \sqrt{3}  +  \sqrt{15}
iii)
( \sqrt{3}  +  \sqrt{2} ) {}^{2}  \\  =  (\sqrt{3})  {}^{2}  +  (\sqrt{2})  {}^{2}  + 2 \times  \sqrt{ 3}  \times  \sqrt{2}  \\  = 3 + 2 + 2 \sqrt{6 }  \\  = 5 + 2 \sqrt{6}
iv)
( \frac{2}{3}  \sqrt{7}  -  \frac{1}{2}  \sqrt{2}  + 6 \sqrt{11} )( \frac{1}{3}  \sqrt{7}  + \frac{3}{2}    \sqrt{2} -  \sqrt{11}) \\  =   \frac{2}{3}  \sqrt{7}  \times  \frac{1}{3}   \sqrt{7}  +  \frac{2}{3}  \sqrt{7}  \times  \frac{3}{2}  \sqrt{2}  -   \frac{2}{3}  \sqrt{7}   \times  \sqrt{11}  \\  -  \frac{1}{2}  \sqrt{2}  \times   \frac{1}{3}  \sqrt{7}  -  \frac{1}{2}  \sqrt{2}   \times  \frac{3}{2}  \sqrt{2}  +  \frac{1}{2}  \sqrt{2}  \times  \sqrt{11}  \\  + 6 \sqrt{11}  \times  \frac{1}{3}  \sqrt{7}  + 6 \sqrt{11}  \times  \frac{3}{2}  \sqrt{2}  - 6 \sqrt{11}  \times  \sqrt{11}  \\  =  \frac{14}{3}  +   \sqrt{14}  -   \frac{2}{3}  \sqrt{77 }  -  \frac{1}{6}  \sqrt{14}  -  \frac{3}{2}  +  \frac{1}{2}  \sqrt{22}  \\  + 2 \sqrt{77}  + 9 \sqrt{22}  - 66 \\  =  \frac{14}{3}  -  \frac{3}{2}  - 66 +  \sqrt{14}   -  \frac{1}{6}  \sqrt{14} -  \frac{2}{3} \sqrt{77}    + 2 \sqrt{77}  \\  +  \frac{1}{2}  \sqrt{22}  + 9 \sqrt{22}  \\  =  \frac{ - 377}{6}  +  \frac{5}{6}  \sqrt{14} +  \frac{4}{3} \sqrt{77 }   +  \frac{19}{2}   \sqrt{22}



HimanshuR: pleazz mark as brainliest
Devchauhanuuu: thik he
HimanshuR: Ohho thanks dude
Devchauhanuuu: koi bhat ni
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