Math, asked by aadarshsharma310798, 9 months ago

If a = √8 - √7 and a = 1/b, then (a^2 + b^2 - 3ab)/(a^2 + ab + b^2) is equal to:​

Answers

Answered by BrainlyPopularman
7

GIVEN :

 \\  \:  \: { \huge{.}} \:  \: { \bold{a =  \sqrt{8} -  \sqrt{7}  }} \\

 \\  \:  \: { \huge{.}} \:  \: { \bold{a =   \dfrac{1}{b}  }} \\

TO FIND :

 \\  \:  \: { \huge{.}} \:  \: { \bold{ \frac{ {a}^{2}  +  {b}^{2} - 3ab }{ {a}^{2} +  {b}^{2} + ab   }  = ?  }} \\

SOLUTION :

• According to the question –

 \\  \:  \implies \:  \: { \bold{ a =  \dfrac{1}{b} }} \\

 \\  \:  \implies \:  \: { \bold{ b =  \dfrac{1}{a} }} \\

 \\  \:  \implies \:  \: { \bold{ b =  \dfrac{1}{ \sqrt{8}  -  \sqrt{7} } }} \\

• Now rationalization –

 \\  \:  \implies \:  \: { \bold{ b =  \dfrac{1}{ \sqrt{8}  -  \sqrt{7} } \times  \dfrac{ \sqrt{8}  +  \sqrt{7} }{ \sqrt{8}  +  \sqrt{7} }  }} \\

 \\  \:  \implies \:  \: { \bold{ b =   \dfrac{ \sqrt{8}  +  \sqrt{7} }{ (\sqrt{8}  +  \sqrt{7} )(\sqrt{8}  -  \sqrt{7})}  }} \\

 \\  \:  \implies \:  \:  { \bold{ b =   \dfrac{ \sqrt{8}  +  \sqrt{7} }{ {( \sqrt{8} )}^{2} -  {( \sqrt{7}) }^{2}  } \:  \:   [  \:  \: \because \:  \: (a + b)(a - b) =  {a}^{2} -  {b}^{2}  ]}} \\

 \\  \:  \implies \:  \:  { \bold{ b =   \dfrac{ \sqrt{8}  +  \sqrt{7} }{ 8 - 7  }}} \\

 \\  \:  \implies \:  \:  { \bold{ b =   { \sqrt{8}  +  \sqrt{7} }}} \\

• Now –

 \\  \:  \:  = \:  \: { \bold{ \frac{ {a}^{2}  +  {b}^{2} - 3ab }{ {a}^{2} +  {b}^{2} + ab   }   }} \\

 \\  \:  \:  = \:  \: { \bold{ \frac{ {a}^{2}  +  {b}^{2} - 2ab - ab }{ {a}^{2} +  {b}^{2} + 2ab   - ab }   }} \\

 \\  \:  \:  = \:  \: { \bold{ \frac{  {(a - b)}^{2} - ab }{  {(a + b)}^{2}   - ab }   \:  \:  \:  \:  \:  \: [   \:  \: \because \:  \: {(a  \pm b)}^{2}  =  {a}^{2}  +  {b}^{2}   \pm 2ab \: ]}} \\

• Now put the values –

 \\  \:  \:  = \:  \: { \bold{ \frac{  {( \sqrt{8}  -  \sqrt{7}  -  \sqrt{8} -  \sqrt{7}  )}^{2} - ( \dfrac{1}{ \cancel b})  \cancel b }{  {( \sqrt{8}  -  \sqrt{7} +  \sqrt{8}  +  \sqrt{7}  )}^{2}   - ( \dfrac{1}{ \cancel b})  \cancel b  }  }} \\

 \\  \:  \:  = \:  \: { \bold{ \frac{  {(  - 2 \sqrt{7}  )}^{2}  - 1}{  {(2 \sqrt{8} ) }^{2} - 1}   }} \\

 \\  \:  \:  = \:  \: { \bold{ \frac{  28  - 1}{  32- 1}   }} \\

 \\  \:  \:  = \:  \: { \bold{ \frac{  27}{  31}   }} \\

• Hence ,  \:  \: { \bold{ \frac{ {a}^{2}  +  {b}^{2} - 3ab }{ {a}^{2} +  {b}^{2} + ab   }  =  \dfrac{27}{31}  }} \\

 \\ \rule{220}{2} \\

Similar questions