Question

几道初二数学题求解

请点击查看大图

Answer 1

你好。
1、36^7-6^12
=(6²)^7-6^12
=6^14-6^12
=6^12(6^2-1)
=6^12×35
=6^10×1260

因为1260能被140整除，所以36^7-6^12能被140整除。

2、x(x+1)(x+2)(x+3)+1
=[x(x+3)][(x+1)(x+2)]+1
=[x²+3x][x²+3x+2]+1
=(x²+3x)(x²+3x)+(x²+3x)*2+1
=(x²+3x)²+2(x²+3x)+1
=(x²+3x+1)²

3、x²+6x+10
=x²+6x+9+1
=(x+3)²+1
因为x²≥0,所以(x+3)²≥0
所以(x+3)²+1≥1,恒为正数.
4、1-1/2²=(1+1/2)(1-1/2)=3/4
(1-1/2²)(1-1/3²)=(1+1/2)(1-1/2)(1+1/3)(1-1/3)
=(3/2)(1/2)(4/3)(2/3)=(1/2)(4/3)=2/3
(1-1/2²)(1-1/3²)(1-1/4²)=(1+1/2)(1-1/2)(1+1/3)(1-1/3)*(1+1/4)(1-1/5)
=(3/2)(1/2)(4/3)(2/3)(5/4)(3/4)=(1/2)(5/4)=5/8
(1-1/2²)(1-1/3²)(1-1/4²).............(1-1/n²)=(1/2)[(n+1)/n]=(n+1)/2n

Answer 2

你好。
1、36^7-6^12
=(6²)^7-6^12
=6^14-6^12
=6^12(6^2-1)
=6^12×35
=6^10×1260

因为1260能被140整除，所以36^7-6^12能被140整除。

2、x(x+1)(x+2)(x+3)+1
=[x(x+3)][(x+1)(x+2)]+1
=[x²+3x][x²+3x+2]+1
=(x²+3x)(x²+3x)+(x²+3x)*2+1
=(x²+3x)²+2(x²+3x)+1
=(x²+3x+1)²

3、x²+6x+10
=x²+6x+9+1
=(x+3)²+1
因为x²≥0,所以(x+3)²≥0
所以(x+3)²+1≥1,恒为正数.

4、1-1/2²=(1+1/2)(1-1/2)=3/4
(1-1/2²)(1-1/3²)=(1+1/2)(1-1/2)(1+1/3)(1-1/3)
=(3/2)(1/2)(4/3)(2/3)=(1/2)(4/3)=2/3