铁假面
幼苗
共回答了17个问题采纳率:76.5% 举报
∵1/(x^2+x)=1/[x(x+1)]=1/x-1/(x+1),
1/(x^2+3x+2)=1/[(x+1)(x+2)]=1/(x+1)-1/(x+2),
1/(x^2+5x+6)=1/[(x+2)(x+3)]=1/(x+2)-1/(x+3),
1/(x^2+7x+12)=1/[(x+3)(x+4)]=1/(x+3)-1/(x+4),
1/(x^2+9x+20)=1/[(x+4)(x+5)]=1/(x+4)-1/(x+5).
将上述5个式子相加,得:
1/(x^2+x)+1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+9x+20)
=1/x-1/(x+5)=(x+5-x)/[x(x+5)]=5/(x^2+5x),
∴原方程可变成:5/(x^2+5x)=5/(x^2+11x-7),∴x^2+5x=x^2+11x-7,
∴6x=7,∴x=7/6.
经检验,x=7/6满足原方程.
∴原方程的解是:x=7/6.
1年前
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