國二數學~~有關於乘法公式和多項式

2011-10-05 4:31 am
1.
計算 1930^2-2*1931*1933+1932*1936+2
用乘法公式解題~~

2.
若a+b=3,ab=1,則a^4+b^4=?

3.
已知a=b+3,則a^2-a-2ab+b+b^2 =?

4.
若展開(x-1)(2x+a)(bx+4)=-2x^3+7x^2+cx-12,則a+b+c=?

回答 (1)

2011-10-05 5:19 am
✔ 最佳答案

1930^2-2*1931*1933+1932*1936+2
=(1931-1)^2-2*1931*1933+1932*(1932+4)+2
=1931^2-2(1931)+1-3866*1931+1932^2+1932*4+2
=1931(1931-2-3866)+1+1932(1932+4)+2
=1931*(-1937)+3+1936*1932
=-(1931*1937)+3+1931*1936+1936
=(1931)(1936-1937)+1939
=-1931+1939
=8
a^4+b^4
=a^4+2a^2b^2+b^4-2a^b^2
=(a^2+b^2)^2-2a^2b^2
=(a^2+2ab+b^2-2ab)^2-2a^2b^2
=((a+b)^2-2ab)^2-2a^2b^2
=(3^2-2(1))^2-2(1)^2
=(9-2)^2-2
=49-2
=47

a^2-a-2ab+b+b^2
=(a-b)^2-a+b
=(b+3-b)^2-(b+3)+b
=9-3
=6

lhs=(x-1)(2x+a)(bx+4)
=2bx^3+x^2(ab-2b+8)+x(-ab+4a-8)-4a
rhs=-2x^3+7x^2+cx-12
b=-1
a*-1+2*1+8=7
-a+10=7
a=3
-ab+4a-8=c
-(3)(-1)+4(3)-8=c
3+12-8=c
c=7
參考: Google 計算機的更多資料


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