Question

Solve this anyone please
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Answer 1

Answer:

this is your answer. Hope you are happy

Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Concept}}}}$

Here in this question, concept of rationalisation is used. We see that we are given a fraction with denominator as irrational number. So firstly we will rationalise it and will compare it with another given equation to find value of a and b.

So let's do it!!

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Given equation is::

$\implies\sf \dfrac{4+\sqrt7}{4-\sqrt 7}=a+b\sqrt 7$

Let's consider LHS!

$\implies\sf \dfrac{4+\sqrt7}{4-\sqrt 7}$

Now multiplying 4+√7 on both numerator and denominator to rationalise it::

$\implies\sf \dfrac{4+\sqrt7}{4-\sqrt 7}\times\bf \Bigg( \dfrac{4+\sqrt 7}{4+\sqrt 7}\Bigg)$

Now multiplying it::

$\implies\sf \dfrac{(4+\sqrt7)\bf (4+\sqrt 7)}{(4-\sqrt 7)\bf (4+\sqrt 7)}$

$\implies\sf \dfrac{(4+\sqrt7)^2}{(4-\sqrt 7)(4+\sqrt 7)}$

Now applying algeberic identities :-

• (A+B)²=A²+B²+2AB

• (A+B)(A-B)=A²-B²

$\implies\sf \dfrac{(4)^2+(\sqrt7)^2+2(4)(\sqrt 7)}{(4)^2-(\sqrt 7)^2}$

Now solving it::

$\implies\sf \dfrac{16+7+8\sqrt 7}{16-7}$

$\implies\sf \dfrac{23+8\sqrt 7}{9}$

$\implies\sf \dfrac{23}{9}+\dfrac{8}{9}\sqrt 7$

By comparing it with RHS, we get::

$\boxed{\sf{\bf{ a}}=\dfrac{23}{9}},\boxed{\sf{\bf{b}}=\dfrac{8}{9}}$

Hence we obtained the values of a and b.