Question

please answer this ​

Answer 1

Answer:

a=0 and b= -2/3

Step-by-step explanation:

First let us understand what rationalising is,

Rationalising is a mathematical procedure in which a denominator of a fraction is multiplied or divided to remove the radical signs.

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Now,with this clarified,let us solve the problem:

Given that,

$\sf{ \frac{ \sqrt{7} - 1 }{ \sqrt{7 + 1} } - \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 } = a + b \sqrt{7} }$

To find out the values of a and b

Now on rationalising independently,

$\sf{ \frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1 } \times \frac{ \sqrt{7} - 1 }{ \sqrt{7} - 1}} \\ \\ = \sf{ \frac{( \sqrt{7} - 1) {}^{2} }{ \sqrt{7 {}^{2} - 1 {}^{2} } } } \\ \\ = \sf{ \frac{7 + 1 - 2 \sqrt{7} }{6} } \\ \\ = \sf{ \frac{8 - 2 \sqrt{7} }{6} = \frac{4 - \sqrt{7} }{3} }$

Second part of the expression,

$\sf{ \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1 }} \\ \\ = \sf{ \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1}} \\ \\ = \sf{\frac{( \sqrt{7} + 1) {}^{2} }{ \sqrt{7 {}^{2} } - 1 {}^{2} }} \\ \\ \sf{ = \frac{8 + 2 \sqrt{7} }{6} = \frac{4 + \sqrt{7} }{3} }$

Replacing with simplified values,we get:

$\sf {\frac{4 - \sqrt{7} }{3} - \frac{4 + \sqrt{7} }{3} } \\ \\ = \sf{ \frac{4 - \sqrt{7} - (4 + \sqrt{7)} }{3} } \\ \\ = \sf{ \frac{ - 2 \sqrt{7} }{3} }$

The values are 0 and -2/3 respectively

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•Identities used:

(a+b)²=a²+2ab+b²

(a-b)²=a²-2ab+b²

(a+b)(a-b)=a² - b²

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