Math, asked by gauri2508, 11 months ago

please tell me how to solve this problem please​

Attachments:

Answers

Answered by Anonymous
4

Answer with Explanation :

\sf (\frac{\sqrt{7}+i\sqrt{3}}{\sqrt{7}-i\sqrt{3}})+(\frac{\sqrt{7}-i\sqrt{3}}{\sqrt{7}+i\sqrt{3}})

taking LCM

 =  (\frac{( \sqrt{7}  + i \sqrt{3} )( \sqrt{7} + i \sqrt{3} ) + ( \sqrt{7}   - i \sqrt{3})(7 - i \sqrt{3} ) }{(7 - i \sqrt{3})(7 +i \sqrt{3}   )} )

 = ( \frac{ {( \sqrt{7}  + i \sqrt{3}) }^{2} + ( \sqrt{7}  - i \sqrt{3}   ) ^{2} }{(  { \sqrt{7} )}^{2}  -  {(i \sqrt{3}) }^{2}  } )

 = ( \frac{  {( \sqrt{7}  ) ^{2}+ 2 \times ( \sqrt{7})(i \sqrt{3} )}   +  ({ \sqrt{3} i} )^{2}  + ( \sqrt{7}  )^{2} - 2 \times ( \sqrt{7} ) (i \sqrt{3} ) +  {(i \sqrt{3} )}^{2} } {7  + 3} )

  =  (\frac{7 - 3 + 7 - 3}{10}  )

 = ( \frac{4 + 4}{10} ) = ( \frac{4}{5} )

{here i = √(-1) , therefore i²=(√-1)²=-1}

Hence (\frac{4}{5})is a real number.


Anonymous: Nice : )
Anonymous: thanks :-)
Answered by pranay0144
22

Step-by-step explanation:

(

7

−i

3

7

+i

3

)+(

7

+i

3

7

−i

3

)

taking LCM

= (\frac{( \sqrt{7} + i \sqrt{3} )( \sqrt{7} + i \sqrt{3} ) + ( \sqrt{7} - i \sqrt{3})(7 - i \sqrt{3} ) }{(7 - i \sqrt{3})(7 +i \sqrt{3} )} )=(

(7−i

3

)(7+i

3

)

(

7

+i

3

)(

7

+i

3

)+(

7

−i

3

)(7−i

3

)

)

= ( \frac{ {( \sqrt{7} + i \sqrt{3}) }^{2} + ( \sqrt{7} - i \sqrt{3} ) ^{2} }{( { \sqrt{7} )}^{2} - {(i \sqrt{3}) }^{2} } )=(

(

7

)

2

−(i

3

)

2

(

7

+i

3

)

2

+(

7

−i

3

)

2

)

= ( \frac{ {( \sqrt{7} ) ^{2}+ 2 \times ( \sqrt{7})(i \sqrt{3} )} + ({ \sqrt{3} i} )^{2} + ( \sqrt{7} )^{2} - 2 \times ( \sqrt{7} ) (i \sqrt{3} ) + {(i \sqrt{3} )}^{2} } {7 + 3} )=(

7+3

(

7

)

2

+2×(

7

)(i

3

)+(

3

i)

2

+(

7

)

2

−2×(

7

)(i

3

)+(i

3

)

2

)

= (\frac{7 - 3 + 7 - 3}{10} )=(

10

7−3+7−3

)

= ( \frac{4 + 4}{10} ) = ( \frac{4}{5} )=(

10

4+4

)=(

5

4

)

{here i = √(-1) , therefore i²=(√-1)²=-1}

Hence (\frac{4}{5})(

5

4

) is a real number.

Similar questions