Question

pls solve it , on a paper not type and answer it will be convenient and helpfull for me and will that as brainelist​

Answer 1

QUESTION:

Simplify by rationalizing the denominator

$\dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \dfrac{4 \sqrt{3}} { \sqrt{6} + \sqrt{2} } - \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

ANSWER:

$2 \sqrt{6} ( \sqrt{2} - \sqrt{3}) \ \ (or)\ \ 4\sqrt{3}-6\sqrt{2}$

GIVEN:

$\dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \dfrac{4 \sqrt{3}} { \sqrt{6} + \sqrt{2} } - \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

TO SIMPLIFY:

$\dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \dfrac{4 \sqrt{3}} { \sqrt{6} + \sqrt{2} } - \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

EXPLANATION:

$\dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \dfrac{4 \sqrt{3}} { \sqrt{6} + \sqrt{2} } - \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

$Take \ \ \dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} }$

Multiply by √6 - √3 on both numerator and denominator.

$\dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } \times \dfrac{\sqrt{6} - \sqrt{3}}{\sqrt{6} - \sqrt{3}}$

$(A+B)(A-B) = A^2 - B^2$

$\dfrac{3 \sqrt{2}(\sqrt{6} - \sqrt{3}) }{ {(\sqrt{6})}^{2} - (\sqrt{3}) ^{2} }$

$\dfrac{3 \sqrt{12} - 3\sqrt{6} }{6 - 3}$

$\dfrac{3 \sqrt{12} - 3\sqrt{6} }{3}$

$\sqrt{12} - \sqrt{6}$

$Take \ \ - \dfrac{4 \sqrt{3}} { \sqrt{6} + \sqrt{2} }$

Multiply by √6 - √2 on both numerator and denominator.

$- \dfrac{4 \sqrt{3}} { \sqrt{6} + \sqrt{2} } \times \dfrac{ \sqrt{6} -\sqrt{2}}{ \sqrt{6} - \sqrt{2}}$

$- \dfrac{4 \sqrt{3} (\sqrt{6} - \sqrt{2})} { (\sqrt{6})^{2} - (\sqrt{2})^{2} }$

$- \dfrac{4(\sqrt{18} - \sqrt{6})} {6 - 2}$

$\dfrac{4( - \sqrt{18} + \sqrt{6})} {4}$

$\sqrt{6} - \sqrt{18}$

$Take \ \ - \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} }$

Multiply by √2 - √3 on both numerator and denominator.

$- \dfrac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}}$

$- \dfrac{ \sqrt{6} (\sqrt{2} - \sqrt{3})}{ (\sqrt{2})^{2} - (\sqrt{3})^{2} }$

$- \dfrac{ (\sqrt{12} - \sqrt{18})}{ 2 - 3 }$

$- \dfrac{ (\sqrt{12} - \sqrt{18})}{ - 1 }$

$\sqrt{12} - \sqrt{18}$

ADD THE THREE TERMS

$\sqrt{12} - \sqrt{6} + \sqrt{6} - \sqrt{18} + \sqrt{12} - \sqrt{18}$

$2 \sqrt{12} - 2 \sqrt{18}$

$2 \sqrt{6} ( \sqrt{2} - \sqrt{3}) \ \ (or)\ \ 4\sqrt{3}-6\sqrt{2}$

Answer 2

ANSWER:

2(2 root 3-3 root 2)