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解方程1/[x(x+3)]+1/[(x+3)(x+6)]+1/[(x+6)(x+9)]=3/(x平方+1)
题目详情
解方程1/[x(x+3)]+1/[(x+3)(x+6)]+1/[(x+6)(x+9)]=3/(x平方+1)
▼优质解答
答案和解析
1/[x(x+3)]+1/[(x+3)(x+6)]+1/[(x+6)(x+9)]=3/(x平方+1)
1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+6)]+1/3[1/(x+6)-1/(x+9)]=3/(x^2+1)
1/3[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]=3/(x^2+1)
1/3[1/x-1/(x+9)]=3/(x^2+1)
1/3(x+9-x)/x(x+9)=3/(x^2+1)
3/x(x+9)=3/(x^2+1)
x(x+9)=x^2+1
x^2+9x=x^2+1
9x=1
x=1/9
检验：X=1/9是方程的根
1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+6)]+1/3[1/(x+6)-1/(x+9)]=3/(x^2+1)
1/3[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]=3/(x^2+1)
1/3[1/x-1/(x+9)]=3/(x^2+1)
1/3(x+9-x)/x(x+9)=3/(x^2+1)
3/x(x+9)=3/(x^2+1)
x(x+9)=x^2+1
x^2+9x=x^2+1
9x=1
x=1/9
检验：X=1/9是方程的根
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