解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/x+1

解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/x+1
解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/x+1

解方程1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)-1/x+1
原式=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)-1/(x+1)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)-1/(x+1)
=1/(x+1)-1/(x+1)
=0

1/(x²+3x+2)+1/(x²+5x+6)=1/(x²+7x+12)-1/(x+1)
1/[(x+1)(x+2)]+1/[(x+2)(x+3)]=1/[(x+3)(x+4)]-1/(x+1)
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)=1/[(x+3)(x+4)]-1/(x+1)
1/(x+1)-1/(x+3)=1/[(...

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1/(x²+3x+2)+1/(x²+5x+6)=1/(x²+7x+12)-1/(x+1)
1/[(x+1)(x+2)]+1/[(x+2)(x+3)]=1/[(x+3)(x+4)]-1/(x+1)
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)=1/[(x+3)(x+4)]-1/(x+1)
1/(x+1)-1/(x+3)=1/[(x+3)(x+4)]-1/(x+1)
去分母，等式两边同乘以(x+1)(x+3)(x+4)
(x+3)(x+4)-(x+1)(x+4)=(x+1)-(x+3)(x+4)
整理，得
x²+8x=-19
x²+8x+16=-3
(x+4)²=-3
平方项恒非负，方程无解。

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