计算1/1*2+1/2*31/3*4+1/4*5+1/5*6-查字典问答网
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  计算1/1*2+1/2*31/3*4+1/4*5+1/5*6计算1/1*3+1/3*5+1/5*7+1/7*9+1/9*11化简1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)+1/(x+9)(x+12)

  计算1/1*2+1/2*31/3*4+1/4*5+1/5*6

  计算1/1*3+1/3*5+1/5*7+1/7*9+1/9*11

  化简1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)+1/(x+9)(x+12)

1回答
2020-05-14 16:48
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郎志强

  1/(1×2)+1/(2×3)+1/(3×4)+1/(4×5)+1/(5×6)

  做这题之前请看下面一些式子:

  1/(1×2)=1-(1/2)=1/2

  1/(2×3)=(1/2)-(1/3)=1/6

  1/(3×4)=(1/3)-(1/4)=1/12

  …………

  1/(1×2)+1/(2×3)+1/(3×4)+1/(4×5)+1/(5×6)

  原式=1-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+(1/5)-(1/6)

  =1-(1/6)

  =5/6

  1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)

  1/3*5=0.5×[(1/3)-(1/5)]

  1/5*7=0.5×[(1/5-1/7)]……………

  ∴1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)

  原式=0.5[1-(1/3)+(1/3)-(1/5)+(1/5)-……-(1/9)+(1/9)-(1/11)]

  =0.5×[1-(1/11)]

  =0.5×(10/11)

  =5/11

  1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)+1/(x+9)(x+12)

  1/x(x+3)=1/3{(1/x)-[1/(x+3)]}

  1/(x+3)(x+6)=1/3{[1/(x+3)]-1/(x+6)}…………

  1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)+1/(x+9)(x+12)

  原式=1/3{(1/x)-[1/(x+3)]+[1/(x+3)]-[1/(x+6)]+[1/(x+6)]-………-[1/(x+9)]+[1/(x+9)]-[1/(x+11)]}

  =1/3{(1/x)-[1/(x+11)]}

  =1/3[(x+10)/(x²+11x)]

  =(x+10)/(3x²+33x)

2020-05-14 16:53:56

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