中二數學一問(1)

1.(e2+2e+5)+(3e2+7e+6)
= e2+2e+5+3e2+7e+6
= 4e2+9e+11

2.(2h2 - h+9) - (6h2+3h - 4)
= 2h2 - h+9 - 6h2 - 3h+4
= -4h2 - 2h+11

３.(3p2 - 8p - 6) - (5p2 - 5p)
= 3p2 - 8p - 6 - 5p2+5p
= -2p2 - 3p - 6

４.(4m-8mm)+(-2m-3mn)
= 4m-8mm - 2m - 3mn
= 2m - 11mn

5.(-3ac-4bc)+(bc+2ac )
= -3ac-4bc+bc+2ac
= ac - 3ac

6.(3ab+ac+c)-(c-2ac- 3ab)
= 3ab+ac+c - c+2ac+3ab
= 6ab+3ac

7.(-bc+5c - 6b2)-(8c+2b2 - 3abc)
= -bc+5c - 6b2- 8c- 2b2+3abc
= -bc+3abc - 3c - 8b2

8.(7y-4)-(2y2+5y-1)+(3-6y)
= 7y-4 - 2y2- 5y+1+3-6y)
= - 2y2 - 4y

2007-11-06 17:10:30 補充：
只有同項的先可相加或相減a^2和2a^2是同一項

1.(e^2+2e+5)+(3e^2+7e+6)
=e^2+2e+5+3e^2+7e+6
=4e^2+9e+11

2.(2h^2-h+9)-(6h^2+3h-4)
=2h^2-h+9-6h^2-3h+4
= -8h^2-4h+13

3.(3p^2-8p-6)-(5p^2-5p)
=3p^2-8p-6-5p^2+5p
= -2p^2-3p-6

4.(4m-8mn)+(-2m-3mn)
=4m-8mn-2m-3mn
=2m-11mn

5.(-3ac-4bc)+(bc+2ac)
= -3ac-4bc+bc+2ac
= -3bc-ac

6.(3ab+ac+c)-(c-2ac-3ab)
= 3ab+ac+c-c-2ac-3ab
= -ac

7.(-bc+5c-6b^2)-(8c+2b^2-3abc)
= -bc+5c-6b^2-8c-2b^2+3abc
= -8b^2+3abc-bc-3c

8.(7y-4)-(2y^2+5y-1)+(3-6y)
= -2y^2+7y-5y-6y-4+1+3
= -2y^2-4y

1.4e二次＋9e＋11
2.-4h二次-4h+5
3.-2p二次-3p-6
4.2m-11mn
5.-ac-3bc
6.-ac
7.-8b二次-3c-bc-3abc
8.8y-2y二次

1.4e二次＋9ｅ＋11
2.-4h二次＋2h+13
跟住個d自己計，努力！

因式分解?