Question

多项式除以单项式~1,利用乘法公式计算：[(a-b)²+4ab]/(a+b)²2,(a^4-b^4)

Answer 1

1,[(a-b)²+4ab]/(a+b)²

=[a^2-2ab+b^2+4ab]/(a+b)^2

=(a+b)^2/(a+b)^2

=1

2,(a^4-b^4)/(a+b)/(a^2+b^2)

=(a^4-b^4)/[(a+b)*(a^2+b^2)]

=[(a^2-b^2)*(a^2+b^2)]/[(a+b)*(a^2+b^2)]

=(a^2-b^2)/(a+b)

=(a-b)(a+b)/(a+b)

=a-b

3,(6a^n+3+12a^n+2-3a^n)/3a^n-2

=3a^n-2 (2a^5+4a^4-3a^2)/3a^n-2

=2a^5+4a^4-3a^2