1、2001²分之2002²+2002²-22、X的6n+2减去2x的3n+2加上X的平方如果正确追加40分.第一道是计算第二道是因式分解

Question

1、2001²分之2002²+2002²-2
2、X的6n+2减去2x的3n+2加上X的平方
如果正确追加40分.
第一道是计算 第二道是因式分解

Answer 1

2001^2分之2002^2+2002^2-2
=2001^2分之2*2002^2-2
=2001^2分之2(2002^2-1)
=2001^2分之2(2002+1)(2002-1)
=2001分之2*2003
=2001分之4006
x^(6n+2)-2x^(3n+2)+x^2
=x^(6n)*x^2-2x^(3n)*x^2+x^2
=x^2*[x^(6n)-2x^(3n)+1]
=x^2*{[x^(3n)]^2-2*x^(3n)+1}
=x^2*[x^(3n)-1]^2
=x^2*[(x^n)^3-1^3]^2
=x^2*[(x^n-1)(x^(2n)+x^n+1]^2
=x^2*(x^n-1)^2*[x^(2n)+x^n+1]^2.