If not t I = 3 the find the the vaule 1 + 2 о ofn 2²​

Question

If not t
I =
3 the
find the
the vaule
1 +
2
о ofn
2²​

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Mackenzie 1 month 2021-08-18T12:37:22+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-18T12:39:12+00:00

     Hey \:  friend, Here  \: is \:  the  \: answer \:  you  \: were \:  looking  \: for: \: \begin{gathered}a = 2 + \sqrt{3} \\ \\ \frac{1}{a} = \frac{1}{2 + \sqrt{3} } \\\end{gathered} a=2+ 3	 a1	 = 2+ 3	 1	 	  On \:  rationalizing  \: the \:  denominator  \: we  \: get,\begin{gathered}\frac{1}{a} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\\end{gathered} a1	 = 2+ 3	 1	 × 2− 3	 2− 3	 	 	   \: using  \: the  \: identity : (x + y)(x - y) = {x}^{2} - {y}^{2}(x+y)(x−y)=x 2 −y 2  \begin{gathered}\frac{1}{a} = \frac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } \\ \\ \frac{1}{a} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{a} = 2 - \sqrt{3} \\ \\ a - \frac{1}{a}\end{gathered} a1	 = (2) 2 −( 3	 ) 2 2− 3	 	 a1	 = 4−32− 3	 	 a1	 =2− 3	 a− a1 \: 	 	  Putting \:  the \:  values,\begin{gathered}a - \frac{1}{a} = (2 + \sqrt{3} ) - (2 - \sqrt{3} ) \\ \\ a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3} \\ \\ a - \frac{1}{a} = \sqrt{3} + \sqrt{3} \\ \\ a - \frac{1}{a} = 2 \sqrt{3}\end{gathered} a− a1	 =(2+ 3	 )−(2− 3	 )a− a1	 =2+ 3	 −2+ 3	 a− a1	 = 3	 + 3	 a− a1	 =2 3

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    2021-08-18T12:39:12+00:00

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