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    If (√7−1)(√7+1)−(√7+1)(√7−1)\frac{(√7-1)}{(√7+1)}-\frac{(√7+1)}{(√7-1)}(√7+1)(√7−1)​−(√7−1)(√7+1)​  = a + b√7, then the values of a and b are: ​
    Question

    If (71)(7+1)(7+1)(71)\frac{(√7-1)}{(√7+1)}-\frac{(√7+1)}{(√7-1)}  = a + b√7, then the values of a and b are: ​ 

    A.

    a = 0 and b = ((-2))/3

    B.

    a = 0 and b = 0

    C.

    a = - 3/2 and b = √7

    D.

    a = √7 and b = (-2√7)/3

    Correct option is A

    Given:

    (71)(7+1)(7+1)(71)\frac{(√7-1)}{(√7+1)}-\frac{(√7+1)}{(√7-1)}​​ = a + b√7

    Solution:

    (71)(7+1)(7+1)(71)\frac{(√7-1)}{(√7+1)}-\frac{(√7+1)}{(√7-1)}​ = a + b√7

    on rationalization  the above equation

    (71)×71)(7+1)×71)(7+1)×7+1(71×7+1)\frac{(√7-1)\times√7-1)}{(√7+1)\times√7-1)}-\frac{(√7+1)\times√7+1}{(√7-1\times√7+1)}​​ = a + b√7

    82768+276\frac{8-2\sqrt7}{6}-\frac{8+2\sqrt7}{6} =a + b√7

    476\frac{-4\sqrt7}{6} =a + b√7

    so a= 0, b= -2\3


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