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6xy( $\frac{1}{2}x^{2} -\frac{1}{3} xy+\frac{1}{6} y^{2}$

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nschavez123

16.04.2021

Mathematics

$\implies {\blue {\boxed {\boxed {\purple {\sf { 14 }}}}}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

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

$\\= \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } + \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

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

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

$\\ = \frac{7 + 4 \sqrt{3} }{1} + \frac{7 - 4 \sqrt{3} }{1}$

$\\= 7 + 4 \sqrt{3} + (7 - 4 \sqrt{3} )$

$\\= 7 + 4 \sqrt{3} + 7 -4 \sqrt{3}$

$\\= 14$

Note:

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

$\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{♨}}}}}$

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