数学题[分式]

观察等式：1/1*2=1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4

猜想并写出：1/n(n+2)=

探究并解方程1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18

2.计算：(1-1/2²)(1-1/3²)(1-1/4²)……(1-2002²)(1-1/2003²)

3.已知[(x+2/x)-4][(x²)+(4/x²)-m]=0，求m

用写的

1/n(n+2)=(1/n)-(1/n+2)

1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/(2x+18)

3/x(x+3)+3/(x+3)(x+6)+3/(x+6)(x+9)=9/(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/(2x+18)

1/x=11/(2x+18)

11x=2x+18

x=2

太复杂...

x+2/x=4

(x+2/x)²=4+m

m+4=16

m=12

m=12

①1/n(n+2)=1/2[1/n-1/(n+2)] 1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/(2x+18)3/x(x+3)+3/(x+3)(x+6)+3/(x+6)(x+9)=9/(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/(2x+18)1/x=11/(2x+18)11x=2x+18x=2②(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2002²)(1-1/2003²)=（1-1/2）（1+1/2）（1-1/3）（1+1/3）（1-1/4）（1+1/4）......（1-1/2002）（1+1/2002）（1-1/2003）（1+1/2003）=1/2×3/2×2/3×4/3×3/4×......2001/2002×2003/2002×2002/2003×2004/2003=1002/2003③[(x+2/x)-4][(x²)+(4/x²)-m]=0，x+2/x=4,(x²)+(4/x²)=16-4=12,m=12

①1/n(n+2)=(1/2)*(1/n-1/(n+2))

②1/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))=3/2x+18

③原式=(1+1/2)(1-1/2)(1+1/3)(1-1/3)````(1+1/2003)(1-1/2003)=1002/2003

④m=85/9   ((x+2/x=4) 所以x=2/3 所以m=85/9)

1/n*(n+1)=1/n-1/(n+1)

1/n-1/(n+2)

1/n(n+2)=1/2[1/n-1/(n+2)]

1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/(2x+18)

3/x(x+3)+3/(x+3)(x+6)+3/(x+6)(x+9)=9/(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/(2x+18)

1/x=11/(2x+18)

11x=2x+18

x=2

(1-1/2²)(1-1/3²)(1-1/4²)……(1-2002²)(1-1/2003²)

=(1+1/2)(1-1/2)(1+1/3)(1-1/3).......(1+1/2003)(1-1/2003)

=(1+1/2)(1+1/3).......(1+1/2003)(1-1/2)(1-1/3)... (1-1/2003)

=3/2  ×4/3×......2004/2003   ×1/2 ×2/3×.....×2002/2003

=2004/2   ×1/2003

=1002/2003