Question

化简：化简：1／x²-1+1／x²+4x+3+1／x²+8x+15+…+1／x²+4nx+4n²-1

Answer 1

题目应该是1／(x²-1)+1／(x²+4x+3)+1／(x²+8x+15)+…+1／(x²+4nx+4n²-1) 吧？

原式=1/(x+1)(x-1) + 1/(x+1)(x+3) + 1/(x+3)(x+5)+...+1/(x+2n+1)(x+2n-1)
=0.5( 1/(x-1)-1/(x+1)+1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+.....+1/(x+2n-1)-1/(2n+1) )
=0.5 ( 1/(x-1)-1/(x+2n+1) )
后面的你应该会了

原理很简单 就是把每一项拆成两项 然后隔项相消

Answer 2

1/(x²-1)+1/(x²+4x+3)+1/(x²+8x+15)+…+1/(x²+4nx+4n²-1)
=1/(x-1)(x+1) + 1/(x+1)(x+3) + 1/(x+3)(x+5) + ... + 1/(x+2n-1)(x+2n+1)
=1/2 * [1/(x-1) - 1/(x+1) + 1/(x+1) - 1/(x+3) + 1/(x+3) - 1/(x+5) + ... + 1/(x+2n-1) - 1/(x+2n+1)]
中间的各式相加为零
=1/2 * [1/(x-1) - 1/(x+2n+1)]
=1/2 * [(x+2n+1-x+1)/(x-1)(x+2n+1)]
=(n+1)/[(x-1)(x+2n+1)]

Answer 3

原式=1/(x+1)(x-1) + 1/(x+1)(x+3) + 1/(x+3)(x+5)+...+1/(x+2n+1)(x+2n-1)
=0.5( 1/(x-1)-1/(x+1)+1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+.....+1/(x+2n-1)-1/(2n+1) )
=0.5 ( 1/(x-1)-1/(x+2n+1) )