(1)因式分解 x²-2xy+y²-5x+5y-6 (2)用简便运算 (1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2002²)(1-1/2003²)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 01:33:25

(1)因式分解 x²-2xy+y²-5x+5y-6 (2)用简便运算 (1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2002²)(1-1/2003²)
(1)因式分解 x²-2xy+y²-5x+5y-6 (2)用简便运算 (1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2002²)(1-1/2003²)

(1)因式分解 x²-2xy+y²-5x+5y-6 (2)用简便运算 (1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2002²)(1-1/2003²)
(1)因式分解 x²-2xy+y²-5x+5y-6
=(x-y)²-5(x-y)-6
=(x-y-6)(x-y+1)
(2)用简便运算 (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 × …… × 2002/2003 × 2004/2003
=1/2 ×2004/2003
=1002/2003

x^2-2xy+y^2-5x+5y-6=(x^2-2xy+y^2)-5(x-y)-6=(x-y)^2-5(x-y)-6=[(x-y)-6][(x-y)+1]=(x-y-6)(x-y+1)

(x-y-5)(x-y)-6;1002/2003

(1) x²-2xy+y²-5x+5y-6
=(x-y)²-5(x-y)-6
=(x-y-6)(x-y+1)
(2) (1-1/2²)(1-1/3²)(1-1/4²)……(1-1/2002²)(1-1/2003²)
=(1+1/2)(1-...

全部展开

(1) x²-2xy+y²-5x+5y-6
=(x-y)²-5(x-y)-6
=(x-y-6)(x-y+1)
(2) (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)
=(3/2)(1/2)(4/3)(2/3)(5/4) (3/4)(6/5)(4/5)……(2003/2002)(2001/2002)(2004/2003)(2002/2003)
=(1/2)(2004/2003) ^
=1002/2003
1、4项3、6项5、8项相乘均为1.因此可推出7、10项9、12项……相乘均为1.,最后即第二项及倒数第二项保留,相乘即可。

收起