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    The value of (1+(1−2)−1+2−(1−2)−2)2\left( 1 + (1 - 2)^{-1} + 2 - (1 - 2)^{-2} \right)^2(1+(1−2)−1+2−(1−2)−2)2​ is:
    Question

    The value of (1+(12)1+2(12)2)2\left( 1 + (1 - 2)^{-1} + 2 - (1 - 2)^{-2} \right)^2​ is:

    A.

    0

    B.

    9

    C.

    1

    D.

    8

    Correct option is C

    Given:

    (1+(12)1+2(12)2)2\left( 1 + (1 - 2)^{-1} + 2 - (1 - 2)^{-2} \right)^2

    Concept Used:

    am=1ama^{-m} = \frac1 {a^{m}}​​

    Solution:

    =(1+(12)1+2(12)2)2 =(1+(1)1+2(1)2)2 =(1+1(1)1+21(1)2)2 =(11+21)2= \left( 1 + (1 - 2)^{-1} + 2 - (1 - 2)^{-2} \right)^2\\ \ \\=\left( 1 + (-1)^{-1} + 2 - (-1)^{-2} \right)^2\\ \ \\=\left( 1 + \frac1{(-1)^{1}} + 2 - \frac1{(-1)^{2}} \right)^2\\ \ \\=\left( 1 - 1 + 2 - 1 \right)^2

    = 12 

    = 1

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