解方程:(1/x²+3x+2)+(1/x²+5x+6)+(1/x²+7x+12)+(1/x²+9x+20)=1/8
解方程:(1/x²+3x+2)+(1/x²+5x+6)+(1/x²+7x+12)+(1/x²+9x+20)=1/8
数学人气:452 ℃时间:2019-10-20 18:24:52
优质解答
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/81/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=1/81/(x+1)-1/(x+5)=1/8(x+5-x-1)/(x+1)(x+5)=1/832=x²+6x+5x²+6x-27=0(x+9)...为什么是这么写呢?1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=1/8裂项看一个1/(x+1)(x+2)=[(x+2)-(x+1)]/ (x+1)(x+2) =(x+2)/(x+1)(x+2)-(x+1)/(x+1)(x+2) =1/(x+1)-1/(x+2)
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