Math, asked by BhawnaAggarwalBT, 1 year ago


plz do it right now
it is urgent
if you can please do my other questions also from my profile
otherwise report them

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Answers

Answered by abhi569
50

 \bold{a =   \frac{ \sqrt{5} + 1 }{ \sqrt{5}  - 1}  }\\  \\   \bold{ \: by \: rationalization} \\  \\  \\  \\  \bold{a =  \frac{ \sqrt{5}  + 1}{ \sqrt{5}  - 1}  \times  \frac{ \sqrt{5}  + 1}{ \sqrt{5}  + 1}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \: a =  \frac{5 + 1 + 2 \sqrt{5} }{5 - 1}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \:  a =  \frac{6 + 2 \sqrt{5} }{4}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \: a =   \frac{3 +  \sqrt{5} }{2}  } \\  \\  \\  \bold{ \:  {a}^{2}  =   {( \frac{3 +  \sqrt{5}) }{2}) }^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \rightarrow \: {a}^{2}   =  \frac{9 + 5 + 6 \sqrt{5} }{4}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \:  {a}^{2}  =  \: \frac{14 + 6 \sqrt{5} }{4}   \:  \:  \:  \:  \:  \:  \:  \rightarrow \:  {a}^{2}  =   \frac{7 + 3 \sqrt{5} }{2} }\:  \:  \:  \:  \:  \:  \:  \: ...(i)





 \bold{ b =  \frac{ \sqrt{5}  - 1}{ \sqrt{5} + 1 }  \:  \:  \: \:  \:  \:   \:  \:  \:} \\  \\  \bold{ \: by \: rationalization} \\  \\  \\  \\  \bold{b = \frac{ \sqrt{5}  - 1}{ \sqrt{5} + 1  }   \times  \frac{ \sqrt{5}   - 1}{ \sqrt{5} - 1 }  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \: b =  \frac{5 + 1 - 2 \sqrt{5} }{5 - 1} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \:b =   \frac{6 - 2 \sqrt{5} }{4}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \: b =  \frac{3 -  \sqrt{5} }{2} } \\  \\  \\  \bold{ {b}^{2} =  { (\frac{3 -  \sqrt{5} }{2} )}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \:  {b}^{2}   =  \frac{9 + 5 - 6\sqrt{5} }{4} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \rightarrow \:  {b}^{2}  =  \frac{14 - 6  \sqrt{5} }{4}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  {b}^{2}  =  \frac{7 - 3 \sqrt{5} }{2} } \:  \:  \:  \:  \:  \:  \: ...(ii)





 \bold{ \: ab =  \frac{ \sqrt{5}  + 1}{ \sqrt{5} - 1 }  \times  \frac{ \sqrt{5}  - 1}{ \sqrt{5}  + 1} } \\  \\   \\ \bold{ab = 1} \:  \:  \:  \:  \:  \: ....(iii)








 \bold{ \underline{ \: Now  \: }}




  \bold{ \: \frac{ {a}^{2}  + ab +  {b}^{2} }{ {a}^{2} - ab +  {b}^{2}  }} \\  \\  \\  \\    \bold{ \: \: <br />Putting  \: values  \: from ( i ) ,  ( ii ) ,  ( iii ) , } \\  \\  \\  \bold{ \:  =  &gt;  \:  \:  \frac{ \frac{7 + 3 \sqrt{5} }{2}   +  1 +  \frac{7 - 3 \sqrt{5} }{2} }{\frac{7 + 3 \sqrt{5} }{2}    -  1 +  \frac{7 - 3 \sqrt{5} }{2} }  } \\  \\  \\  \\  =  &gt;  \bold{  \frac{ \frac{7 + 3 \sqrt{5}  + 2 + 7 - 3 \sqrt{5}}{2} }{ \frac{7 + 3 \sqrt{5}  - 2 + 7 - 3 \sqrt{5} }{2} }  } \:  \:  \:  \:  \\  \\  \\  \\  =  &gt;   \bold{\frac{ \frac{16}{2} }{ \frac{12}{2} } } \\  \\  \\  \\   =  &gt; \bold{ \: \frac{8}{6}  } \\  \\  \\  \\  =  &gt;  \bold{ \: \frac{4}{3}  }



abhi569: Is the answer correct ?
BhawnaAggarwalBT: yes my answer is also 4/3
abhi569: Ok
abhi569: (-;
jagrutiunagarpd6dai: My answer is aslo correct
akhlaka: Excellent answer bhaiya... :)
abhi569: (-:
Answered by siddhartharao77
40
 Given : a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}

 = &gt; \frac{\sqrt{5} + 1}{\sqrt{5} - 1} * \frac{\sqrt{5} + 1}{\sqrt{5} + 1}

 = &gt; \frac{(\sqrt{5} + 1)^2}{(\sqrt{5)^2 - (1)^2}}

 = &gt; \frac{5 + 1 + 2\sqrt{5}}{4}

 = &gt; \frac{6 + 2\sqrt{5}}{4}

 = &gt; \frac{3 + \sqrt{5}}{2}

Now,

 = &gt; a^2 = (\frac{3 + \sqrt{5}}{2})^2

 = &gt; \frac{9 + 5 + 6\sqrt{5}}{4}

 = &gt; \frac{14 + 6\sqrt{5}}{4}

 = &gt; \frac{7 + 3\sqrt{5}}{2}

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 Given : b = \frac{\sqrt{5} - 1}{\sqrt{5 + 1}}

 = &gt; \frac{\sqrt{5} - 1}{\sqrt{5} + 1} * \frac{\sqrt{5} - 1}{\sqrt{5} - 1}

 = &gt; \frac{(\sqrt{5} - 1)^2}{(\sqrt{5})^2 - (1)^2}

 = &gt; \frac{5 + 1 - 2\sqrt{5}}{5 - 1}

 = &gt; \frac{6 - 2\sqrt{5}}{4}

 = &gt; \frac{3 - \sqrt{5}}{2}

Now,

 = &gt; b^2 = \frac{(3 - \sqrt{5})^2}{2^2}

 = &gt; \frac{9 + 5 - 6\sqrt{5}}{4}

 = &gt; \frac{14 - 6\sqrt{5}}{4}

 = &gt; \frac{7 - 3\sqrt{5}}{2}

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BhawnaAggarwalBT: for how many time you use brainly
BhawnaAggarwalBT: ????
siddhartharao77: I didnt understand your question. is it from how long i am using brainly? (or) In a day how many hours i use brainly?
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