Plz help me in this Q here..I'll mark u as BRAINLEIST!, 9grd Q..
Simplify:
(2√6 / √2+√3) + (6√2 / √6+√3) - (8√3 / √6+√2)
Hope u know that first u have to fractionalize each of them..
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Hey friend,
Here is the answer you were looking for:
\begin{lgathered}\frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } + \frac{6 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \frac{8 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } + \frac{6 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } \times \frac{ \sqrt{6} - \sqrt{3} }{ \sqrt{6} - \sqrt{3} } - \frac{8 \sqrt{3} }{ \sqrt{6 } + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{2 \sqrt{6} \times \sqrt{2} - 2 \sqrt{6} \times \sqrt{3} }{ {( \sqrt{2}) }^{2} - {( \sqrt{3}) }^{2} } + \frac{6 \sqrt{2} \times \sqrt{6} - 6 \sqrt{2} \times \sqrt{3} }{ {( \sqrt{6}) }^{2} - {( \sqrt{3}) }^{2} } - \frac{8 \sqrt{3} \times \sqrt{6} - 8 \sqrt{3} \times \sqrt{2} }{ {( \sqrt{6} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{2 \sqrt{2 \times 2 \times 3} - 2 \sqrt{3 \times 3 \times 2} }{2 - 3} + \frac{6 \sqrt{2 \times 2 \times 3} - 6 \sqrt{6} }{6 - 3} - \frac{8 \sqrt{3 \times 3 \times 2} - 8 \sqrt{6} }{6 - 2} \\ \\ = \frac{2 \times 2 \sqrt{3} - 2 \times 3 \sqrt{2} }{ - 1 } + \frac{6 \times 2 \sqrt{3} - 6 \sqrt{6} }{3} - \frac{8 \times 3 \sqrt{2} - 8 \sqrt{6} }{4} \\ \\ = - 4 \sqrt{3} + 6 \sqrt{2} + \frac{12 \sqrt{3} - 6 \sqrt{6} }{3} - \frac{24 \sqrt{2} - 8 \sqrt{6} }{4} \\ \\ = \frac{ - 4 \sqrt{3} \times 12 + 6 \sqrt{2} \times 12 + 12 \sqrt{3} \times 4 - 6 \sqrt{6} \times 4 - 24 \sqrt{2} \times 3 + 8 \sqrt{6} \times 3}{12} \\ \\ = \frac{ - 48 \sqrt{3} + 72 \sqrt{2} + 48 \sqrt{3} - 24 \sqrt{6} - 72 \sqrt{2} + 24 \sqrt{6} }{12} \\ \\ = \frac{0}{12} \\ \\ = 0\end{lgathered}2+326+6+362−6+283onrationalizingthedenominatorweget=2+326×2−32−3+6+362×6−36−3−6+283×6−26−2usingtheidentity(a+b)(a−b)=a2−b2=(2)2−(3)226×2−26×3+(6)2−(3)262×6−62×3−(6)2−(2)283×6−83×2=2−322×2×3−23×3×2+6−362×2×3−66−6−283×3×2−86=−12×23−2×32+36×23−66−48×32−86=−43+62+3123−66−4242−86=12−43×12+62×12+123×4−66×4−242×3+86×3=12−483+722+483−246−722+246=120=0
Here is the answer you were looking for:
\begin{lgathered}\frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } + \frac{6 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \frac{8 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } + \frac{6 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } \times \frac{ \sqrt{6} - \sqrt{3} }{ \sqrt{6} - \sqrt{3} } - \frac{8 \sqrt{3} }{ \sqrt{6 } + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{2 \sqrt{6} \times \sqrt{2} - 2 \sqrt{6} \times \sqrt{3} }{ {( \sqrt{2}) }^{2} - {( \sqrt{3}) }^{2} } + \frac{6 \sqrt{2} \times \sqrt{6} - 6 \sqrt{2} \times \sqrt{3} }{ {( \sqrt{6}) }^{2} - {( \sqrt{3}) }^{2} } - \frac{8 \sqrt{3} \times \sqrt{6} - 8 \sqrt{3} \times \sqrt{2} }{ {( \sqrt{6} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{2 \sqrt{2 \times 2 \times 3} - 2 \sqrt{3 \times 3 \times 2} }{2 - 3} + \frac{6 \sqrt{2 \times 2 \times 3} - 6 \sqrt{6} }{6 - 3} - \frac{8 \sqrt{3 \times 3 \times 2} - 8 \sqrt{6} }{6 - 2} \\ \\ = \frac{2 \times 2 \sqrt{3} - 2 \times 3 \sqrt{2} }{ - 1 } + \frac{6 \times 2 \sqrt{3} - 6 \sqrt{6} }{3} - \frac{8 \times 3 \sqrt{2} - 8 \sqrt{6} }{4} \\ \\ = - 4 \sqrt{3} + 6 \sqrt{2} + \frac{12 \sqrt{3} - 6 \sqrt{6} }{3} - \frac{24 \sqrt{2} - 8 \sqrt{6} }{4} \\ \\ = \frac{ - 4 \sqrt{3} \times 12 + 6 \sqrt{2} \times 12 + 12 \sqrt{3} \times 4 - 6 \sqrt{6} \times 4 - 24 \sqrt{2} \times 3 + 8 \sqrt{6} \times 3}{12} \\ \\ = \frac{ - 48 \sqrt{3} + 72 \sqrt{2} + 48 \sqrt{3} - 24 \sqrt{6} - 72 \sqrt{2} + 24 \sqrt{6} }{12} \\ \\ = \frac{0}{12} \\ \\ = 0\end{lgathered}2+326+6+362−6+283onrationalizingthedenominatorweget=2+326×2−32−3+6+362×6−36−3−6+283×6−26−2usingtheidentity(a+b)(a−b)=a2−b2=(2)2−(3)226×2−26×3+(6)2−(3)262×6−62×3−(6)2−(2)283×6−83×2=2−322×2×3−23×3×2+6−362×2×3−66−6−283×3×2−86=−12×23−2×32+36×23−66−48×32−86=−43+62+3123−66−4242−86=12−43×12+62×12+123×4−66×4−242×3+86×3=12−483+722+483−246−722+246=120=0
yoyo23xtto:
Ooh plz do that for me..
Final and is not wat I want...I need to write it..so indeed its explanation
Thnx
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