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1/x²+x+1/x²+3x+2+1/x²+5x+6+1/x²+7x+12,分式方程,
人气:411 ℃ 时间:2019-08-21 16:20:49
解答
原式=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/x-1/(x+4)
=4/(x²+4x)
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