Question

find the value of A and B if root 7 minus 1 by root 7 + 1 - root 7 + 1 bfind the values of A and B if root 7 minus 1 by root 7 + 1 - root 7 + 1 by root 7 minus 1 is equals to a + b root 7 and (√)root is only with 7 not with one (√)​

Answer 1

Answer:

Hey mate....❤❤

Here is your answer.....

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

Rationalise the denominator...

$( \frac{ \sqrt{7} - 1}{ \sqrt{7} + 1} \times \frac{ \sqrt{7} - 1}{ \sqrt{7} - 1} ) - ( \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1 } ) \\ \\ ( \frac{7 + 1 - 2 \sqrt{7} }{ { \sqrt{7} }^{2} - {1}^{2} } ) - ( \frac{7 + 1 + 2 \sqrt{7} }{ { \sqrt{7} }^{2} - {1}^{2} } ) \\ \\( \frac{8 - 2 \sqrt{7} }{6} ) - ( \frac{8 + 2 \sqrt{7} }{6} ) \\ \\ \frac{8 - 2 \sqrt{7} - (8 + 2 \sqrt{7}) }{6} \\ \\ \frac{8 - 2 \sqrt{7 } - 8 - 2 \sqrt{7} }{6} \\ \frac{ - 4 \sqrt{7} }{6} \\ \\ \frac{ - 2 \sqrt{7} }{3} = a + b \sqrt{7} \\ \\ 0 + \frac{ - 2}{3} \sqrt{7} = a + b \sqrt{7} \\ \\ by \: comparing \: both \: sides \\ \\ a = 0 \\ \\ b = - \frac{ - 2}{3}$

Please mark it as the brainliest answer

Answer 2

Answer:

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