Question

- rationalising
the denominator
1+√2÷2-√2​

Answer 1

AnsweR :-

$\dfrac{1 + \sqrt{2} }{2 - \sqrt{2} }$

$\dfrac{1 + \sqrt{2}}{2 - \sqrt{2} } \: \: \times \dfrac{2 + \sqrt{2} }{2 + \sqrt{2} }$

$\dfrac{2 + 2}{ {(2)}^{2} - \sqrt{({2)}^{2} } }$

$\dfrac{4}{4 - 2}$

$\dfrac{4}{2}$

$\dfrac{2}{1}$

Procedure :-

→ As per the required question we need to rationalise the denominator. After multiplying the denominator with 1+√2/ 2+√2 we get 2+2/(2)²-(√2)².

→ We will put the negetive sign in the denominator because we have used an Algerbric identity that (a+b)(a-b) = a² - b². After that will we get the fraction 4/4-2.

→ Then we will Supract 2 from 4 in the denominator and then we will cancle 4/2 and Hence, the final answer is 2/1.

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Answer 2

$\dfrac{1 + \sqrt{2} }{2 - \sqrt{2} }$

→$\dfrac{1 + \sqrt{2}}{2 - \sqrt{2} } \times \: \dfrac{2 + \sqrt{2} }{2 + \sqrt{2} }$

→$\dfrac{2 + 2}{ {(2)}^{2} - {( \sqrt{2)}}^{2} }$

→ $\dfrac{4}{4 - 2}$

→ $\dfrac{4}{2}$

→$\dfrac{2}{1}$

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