Correct option is CGiven:(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 Concept Used: a−m=1ama^{-m} = \frac1 {a^{m}}a−m=am1Solution:=(1+(1−2)−1+2−(1−2)−2)2 =(1+(−1)−1+2−(−1)−2)2 =(1+1(−1)1+2−1(−1)2)2 =(1−1+2−1)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=(1+(1−2)−1+2−(1−2)−2)2 =(1+(−1)−1+2−(−1)−2)2 =(1+(−1)11+2−(−1)21)2 =(1−1+2−1)2= 12 = 1