Question

simplify
$\frac{2 + \sqrt{5} }{2 - \sqrt{5} } + \frac{2 - \sqrt{5} }{2 + \sqrt{5} } + \frac{3 - \sqrt{8} }{3 + \sqrt{8} }$
can anyone solve this problem​

Answer 1

Answer:

= - (1+6√8)

Step-by-step explanation:

$=> \frac{2+\sqrt{5} }{2-\sqrt{5} } \\\\rationalising \ factor =2+\sqrt{5} \\\\=> (\frac{2+\sqrt{5} }{2-\sqrt{5} }) ( \frac{2+\sqrt{5} }{2+\sqrt{5} }) \\\\=> \frac{(2+\sqrt{5})^{2} }{2^{2}-\sqrt{5} ^{2}} \ \ \ [ (a+b)(a-b)=a^{2}-b^{2}] \\\\=> \frac{4+2(2)(\sqrt{5}) +\sqrt{5} ^{2} }{4-5} \\\\=> \frac{4+5+4\sqrt{5} }{-1} \\\\=> -(9+4\sqrt{5})$

$=> \frac{2-\sqrt{5} }{2+\sqrt{5} } \\\\rationalising \ factor =2-\sqrt{5} \\\\=> (\frac{2-\sqrt{5} }{2+\sqrt{5} }) ( \frac{2-\sqrt{5} }{2-\sqrt{5} }) \\\\=> \frac{(2-\sqrt{5})^{2} }{2^{2}-\sqrt{5} ^{2}} \ \ \ [ (a+b)(a-b)=a^{2}-b^{2}] \\\\=> \frac{4-2(2)(\sqrt{5}) +\sqrt{5} ^{2} }{4-5} \\\\=> \frac{4+5-4\sqrt{5} }{-1} \\\\=> -(9-4\sqrt{5})$

$=>4\sqrt{5} -9$

$=>\frac{3-\sqrt{8}}{3+\sqrt{8}} \\\\ rationalising \ factor=3-\sqrt{8} \\\\=> (\frac{3-\sqrt{8}}{3+\sqrt{8}})(\frac{3-\sqrt{8}}{3-\sqrt{8}}) \\\\=> \frac{(3-\sqrt{8})^{2}}{3^{2}- \sqrt{8} ^{2}} \ \ \ [(a+b)(a-b)=a^{2}-b^{2}] \\\\ => \frac{9+8-2(3)(\sqrt{8})}{9-8} \\\\=>\frac{17-6\sqrt{8}}{1} \\\\=> 17-6\sqrt{8}$

$adding \ all \ these, \\ = -(9+4\sqrt{5})+4\sqrt{5}-9 +17-6\sqrt{8} \\= -9-4\sqrt{5}+4\sqrt{5}-9+17-6\sqrt{8}\\=-18+17-6\sqrt{8} \\=-1-6\sqrt{8} \\=-(1+6\sqrt{8}) \\\\ HOPE \ YOU \ GOT \ IT..!$

Answer 2

ANSWER:

$\tt \dfrac{2 + \sqrt{5} }{2 - \sqrt{5} } + \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} } + \dfrac{3 - \sqrt{8} }{3 + \sqrt{8} }= - 1 -12\sqrt{2}$

GIVEN:

$\tt \dfrac{2 + \sqrt{5} }{2 - \sqrt{5} } + \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} } + \dfrac{3 - \sqrt{8} }{3 + \sqrt{8} }$

TO SIMPLIFY:

$\tt \dfrac{2 + \sqrt{5} }{2 - \sqrt{5} } + \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} } + \dfrac{3 - \sqrt{8} }{3 + \sqrt{8} }$

EXPLANATION:

$\tt Let \ D =\dfrac{2 + \sqrt{5} }{2 - \sqrt{5} } + \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} } + \dfrac{3 - \sqrt{8} }{3 + \sqrt{8} }$

$\tt Let \ A = \dfrac{2 + \sqrt{5} }{2 - \sqrt{5} }$

$\tt Let\ B = \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} }$

$\tt Let \ C=\dfrac{3 - \sqrt{8} }{3 + \sqrt{8} }$

$\boxed{\bold{\large{\gray{A + B + C = D}}}}$

Take A:

$\sf Multiply\ and\ divide\ by\ 2 + \sqrt{5}$

$\tt \leadsto \dfrac{2 + \sqrt{5} }{2 - \sqrt{5} } \times \dfrac{2 + \sqrt{5} }{2 + \sqrt{5} }$

$\boxed{\bold{\large{\gray{(A + B)(A+ B)=(A+B)^2}}}}$

$\boxed{\bold{\large{\gray{(A + B)(A - B)=A ^{2} - B^2}}}}$

$\tt \leadsto \dfrac{(2 + \sqrt{5} ) ^{2} }{ {2}^{2} - (\sqrt{5})^{2} }$

$\boxed{\bold{\large{\gray{(A + B)^2 = A^2 + 2AB + B^2}}}}$

$\tt \leadsto \dfrac{{2}^{2} + 4 \sqrt{5} + (\sqrt{5})^{2} }{4 - 5 }$

$\tt \leadsto \dfrac{4 + 4 \sqrt{5} + 5 }{ - 1}$

$\tt \leadsto - 9 - 4 \sqrt{5}$

Take B:

$\sf Multiply\ and\ divide\ by\ 2 - \sqrt{5}$

$\tt \leadsto \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} } \times \dfrac{2 - \sqrt{5} }{2 - \sqrt{5} }$

$\boxed{\bold{\large{\gray{(A - B)(A - B)=(A - B)^2}}}}$

$\boxed{\bold{\large{\gray{(A + B)(A - B)=A ^{2} - B^2}}}}$

$\tt \leadsto \dfrac{(2 - \sqrt{5} ) ^{2} }{ {2}^{2} - (\sqrt{5})^{2} }$

$\boxed{\bold{\large{\gray{(A - B)^2 = A^2 - 2AB + B^2}}}}$

$\tt \leadsto \dfrac{{2}^{2} - 4 \sqrt{5} + (\sqrt{5})^{2} }{4 - 5 }$

$\tt \leadsto \dfrac{4 - 4\sqrt{5} + 5 }{ - 1 }$

$\tt \leadsto - 9 + 4 \sqrt{5}$

Take C:

$\sf Multiply\ and\ divide \ by \ 3 - \sqrt{8}$

$\tt \leadsto\dfrac{3 - \sqrt{8} }{3 + \sqrt{8} } \times \dfrac{3 - \sqrt{8} }{3 - \sqrt{8} }$

$\boxed{\bold{\large{\gray{(A - B)(A - B)=(A - B)^2}}}}$

$\boxed{\bold{\large{\gray{(A + B)(A - B)=A ^{2} - B^2}}}}$

$\tt \leadsto\dfrac{(3 - \sqrt{8} )^{2} }{{3}^{2} - (\sqrt{8}) ^{2} }$

$\boxed{\bold{\large{\gray{(A - B)^2 = A^2 - 2AB + B^2}}}}$

$\tt \leadsto\dfrac{9 -6 \sqrt{8} + 8 }{9 - 8}$

$\tt \leadsto\dfrac{17 -6 \sqrt{8} }{1}$

$\tt \leadsto17 -6 \sqrt{8}$

Now add A, B and C [A + B + C = D]

$\tt \leadsto - 9 - 4 \sqrt{5} - 9 + 4 \sqrt{5} + 17 -6 \sqrt{8}$

$\tt \leadsto - 18 + 17 -6 \sqrt{8}$

$\tt \leadsto - 1 -6 \sqrt{8}$

$\tt \leadsto - 1 -6 \sqrt{4\times2}$

$\tt \leadsto - 1 - 6(2) \sqrt{2}$

$\tt \leadsto - 1 - 12 \sqrt{2}$