作业帮解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/(x+4)=0
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解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/(x+4)=0
解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/(x+4)=0
解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/(x+4)=0
原方程可化为：
1/[(x+1)(x+2)] + 1/[(x+2)(x+3)] +1/[(x+3)(x+4)] - 1/(x+4)=0
裂项得：
1/(x+1) - 1/(x+2) + 1/(x+2) - 1/(x+3) + 1/(x+3) - 1/(x+4) - 1/(x+4)=0
即：1/(x+1) - 2/(x+4)=0
1/(x+1) = 2/(x+4)
x+4=2x+2
解得：x=2
经检验,原方程的解为：x=2