计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/05 14:54:44
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?计算1/
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?
1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]
=1/3[1/x-1/(x+3)]+[1/(x+3)-1/(x+6)]+.+[1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+3)+1/(x+3)-1/(x+6)+.+1/(x+15)-1/(x+18)]
=1/x-1/(x+18)
=1/3*(x+18-x)/x(x+18)
=6/(x^2+18x)
1/[x(x+3)]+1/[(x+3)(x+6)]+......+1/[(x+15)(x+18)]
={3/[x(x+3)]+3/[(x+3)(x+6)]+......+3/[(x+15)(x+18)]}/3
={[(1/x-1/(x+3)]+[1/(x+3)-1/(x+6)]+......+[1/(x+15)-1/(x+18)]}/3
=[1/x-1/(x+18)]/3
=[(x+18-x)/(x(x+18)]/3
=6/(x^2+18x)
用裂项法
原式=1/3*【1/x-1/（x+3）+1/（x+3）-1/（x+6）+...1/（x+15）-1/（x+18）】
中括号里的前后消掉，
原式=1/3*【1/x-1/（x+18）】
=-6/【x（x+18)]
计算:x/(1-x)-1/(x-3)
计算:|3-x|+|x-1|(x>5)
计算：(3/x)+(x/x²-1)-(4/1-x)
计算(x+1)(x+2)(x+3)(x+4)
计算：（x+1)(x-2)(x-3)(x-6)
计算 x(x+3)-(x-1)(x-2）
如何计算 (x+1)(x+2)(x+3)(x+4)
计算(X-1)(X-2)(X-3)(X-4)
(x-1)(x-2)(x-3)(x-4)-24计算
计算：x/(x-1)-(x+3)/(x^-1)*(x^2+2x+1)/(x+3)
计算:[(x^2 - 4)/(x^2 - x - 6) + (x+2)/(x-3)] / (x+1)/(x-3)
计算：1-2x+x^2/x-x^3*xy+x^2y/x-x^2
计算：(2x+3)(3x-4)>(x+1)(x-1)+5x(x+3)+7
2x/x-1+1/x+3-3x+5/x^2+2x-3计算,
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?
计算：1/x(x+3)+1/(x+3)(x+6)+...+1/(x+2004)(x+2007)
计算3x^3-(x^2 +2x)(3x-1)-3(x+5)(x-2)
计算:1/(x+3)(x+4)+1/(x+4)(x+5).1/(x+99)(x+100)