1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/2X+18 规律是1/X(X+3)=1/X-1
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1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/2X+18 规律是1/X(X+3)=1/X-1/(X+3)
分母相差几就是几么 ,
分母相差几就是几么 ,
∵1/X-1/(X+N)
=(X+N)/[X(X+N)]-X/[X(X+N)]
=N/[X(X+N)]
∴1/[X(X+N)]=1/N[1/X-1/(X+N)]
再问: 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/(2X+18) 解出来这个题,100财富给你
再答: ������Ǵ���ģ� 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9) =1/3[1/X-1/(X+3)+1/(X+3)-1/(X+6)+1/(X+6)-1/(X+9)] =1/3[1/X-1/(X+9)] =1/3*9/[X(X+9)] =3/(X²+9x)
再问: ���ǽⷽ�̣�ͬ־����������������
再答: ����һ���䡣ֱ��˵�ⷽ�̵��ˣ� �����Ϲ��ɵ�X²+9x=2X+18 ��x²+7X-18=0 ���x1=2�� x2=-9(�������⣬��ȥ) ��X=2
再问: ��д����ϸ���ô����ѧϰ����ܺã�����̫������̡�
再答: 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/(2X+18) 1/3[1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)]=3/(2X+18) [1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]=9/(2x+18) 1/x-1/(x+9)=9/2(x+9) (x+9-x)/x(x+9)=9/2(x+9) 9/x(x+9)=9/2(x+9) ∴x(x+9)=2(x+9) x²+9x-2x-18=0 x²+7x-18=0 ∴x+9=0 x-2=0 x=-9 (不符题意,舍去) ∴x=2 若还有疑问,可在HI里复制提问,免得浪费财富。
=(X+N)/[X(X+N)]-X/[X(X+N)]
=N/[X(X+N)]
∴1/[X(X+N)]=1/N[1/X-1/(X+N)]
再问: 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/(2X+18) 解出来这个题,100财富给你
再答: ������Ǵ���ģ� 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9) =1/3[1/X-1/(X+3)+1/(X+3)-1/(X+6)+1/(X+6)-1/(X+9)] =1/3[1/X-1/(X+9)] =1/3*9/[X(X+9)] =3/(X²+9x)
再问: ���ǽⷽ�̣�ͬ־����������������
再答: ����һ���䡣ֱ��˵�ⷽ�̵��ˣ� �����Ϲ��ɵ�X²+9x=2X+18 ��x²+7X-18=0 ���x1=2�� x2=-9(�������⣬��ȥ) ��X=2
再问: ��д����ϸ���ô����ѧϰ����ܺã�����̫������̡�
再答: 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/(2X+18) 1/3[1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)]=3/(2X+18) [1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]=9/(2x+18) 1/x-1/(x+9)=9/2(x+9) (x+9-x)/x(x+9)=9/2(x+9) 9/x(x+9)=9/2(x+9) ∴x(x+9)=2(x+9) x²+9x-2x-18=0 x²+7x-18=0 ∴x+9=0 x-2=0 x=-9 (不符题意,舍去) ∴x=2 若还有疑问,可在HI里复制提问,免得浪费财富。
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