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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=3\2

分类: 作业答案 • 2021-12-20 10:21:39
问题描述：

解方程1\x^2+3x+2+1\x^2+5x+6+1\x^2+7x+12+1\x+4=3\2
解方程1\x^+3x+2 + 1\x^2+5x+6 + 1\x^2+7x+12 + 1\x+4=3\2

答

1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)=3/2
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)=3/2
1/(x+1)=3/2
x+1=2/3
x=-1/3

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