Question

化简x³-1/x³+2x²+2x+1+x³+1/x³-2x²+2x-1-2x²+2/x²-1

当n为正整数时，1/2+1/4+1/8+...+1/2的n次方=？

Answer 1

(x³-1)/(x³+2x²+2x+1)+(x³+1)/(x³-2x²+2x-1)-(2x²+2)/(x²-1)
=(x-1)(x²+x+1)/[(x³+1)+2x(x+1)]+(x+1)(x²-x+1)/[(x³-1)-2x(x-1)]-(2x²+2)/(x²-1)
=(x-1)(x²+x+1)/[(x+1)(x²-x+1)+2x(x+1)]+(x+1)(x²-x+1)/[(x-1)(x²+x+1)-2x(x-1)]-(2x²+2)/(x²-1)
=(x-1)(x²+x+1)/[(x+1)(x²-x+1+2x)]+(x+1)(x²-x+1)/[(x-1)(x²+x+1-2x)]-(2x²+2)/(x²-1)
=(x-1)(x²+x+1)/[(x+1)(x²+x+1)]+(x+1)(x²-x+1)/[(x-1)(x²-x+1)]-(2x²+2)/(x²-1)
=(x-1)/(x+1)+(x+1)/(x-1)-(2x²+2)/(x²-1)
=(x-1)²/(x+1)(x-1)+(x+1)²/(x+1)(x-1)-(2x²+2)/(x²-1)
=(x-1)²/(x²-1)+(x+1)²/(x²-1)-(2x²+2)/(x²-1)
=[(x-1)²+(x+1)²-(2x²+2)]/(x²-1)
=[x²+2x+1+x²-2x+1-2x²-2]/(x²-1)
=0/(x²-1)
=0

令s=1/2+1/4+1/8+...+1/2^n
s/2=1/4+1/8+...+1/2^(n+1)
s-s/2=1/2-1/2^(n+1)
s/2=1/2-1/2^(n+1)
s=1-1/2^n
即1/2+1/4+1/8+...+1/2^n=1-1/2^n

Answer 2

2x^3-2x^2+4x-1+1/x^2