Question

高中指数运算

我指数运算学的不好 请教一下这道题怎么算的？ （图2和图3画横线的部分）谢谢！

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Answer 1

n×2的n次方 -（n-1）×2的n次方
=n×2的n次方-n×2的n次方+1×2的n次方
=0+1×2的n次方
=2的n次方

Answer 2

(a^2+b^2- a^(-2) -b^(-2))/(a^2b^2- a^(-2)b^(-2)) +(a- a^(-1))(b-b^(-1))/(ab+a^(-1)b^(-1)) =(a^4b^2+a^2b^4- b^2 -a^2)/(a^4b^4-1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =[a^2b^2(a^2+b^2)-(a^2+b^2)]/(a^2b^2+1)(a^2b^2-1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =(a^2b^2-1)(a^2+b^2)/(a^2b^2+1)(a^2b^2-1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =(a^2+b^2)/(a^2b^2+1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =[(a^2+b^2)+(a^2- 1)(b^2-1)]/(a^2b^2+1) =(a^2+b^2+a^2b^2-a^2-b^2+1)/(a^2b^2+1) =(a^2b^2+1)/(a^2b^2+1) =1

追问

你写的是什么啊

Answer 3

(a^2+b^2- a^(-2) -b^(-2))/(a^2b^2- a^(-2)b^(-2)) +(a- a^(-1))(b-b^(-1))/(ab+a^(-1)b^(-1)) =(a^4b^2+a^2b^4- b^2 -a^2)/(a^4b^4-1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =[a^2b^2(a^2+b^2)-(a^2+b^2)]/(a^2b^2+1)(a^2b^2-1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =(a^2b^2-1)(a^2+b^2)/(a^2b^2+1)(a^2b^2-1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =(a^2+b^2)/(a^2b^2+1) +(a^2- 1)(b^2-1)/(a^2b^2+1) =[(a^2+b^2)+(a^2- 1)(b^2-1)]/(a^2b^2+1) =(a^2+b^2+a^2b^2-a^2-b^2+1)/(a^2b^2+1) =(a^2b^2+1)/(a^2b^2+1) =1