f.2數 (20) 急

1.
(x+1)+P(x+1)=x+Q
左方=(x+1)+P(x+1)
=(x+2x+1)+(Px+P)
=x+(P+2)x+(P+1)
比較左方和右方的同類項,可得
P+2=0_________________(1)
P+1=Q_________________(2)
從(1),可得:
P=-2
把P=-2代入(2)
-2+1=Q
Q=-1

2.
a(-2)-b(1)+8=0
-2a-b+8=0___________________(1)
b(-2)+a(1)+1=0
a-2b+1=0____________________(2)
解(1)和(2),可得
a=3,b=2

1.
(x+1)^2+P(x+1)=x^2+Q
LS: x^2+2x+1 + Px + p
=x^2+(2+p)x+(1+p)
so x^2+(2+p)x+(1+p)=x^2+0x+Q
so [同樣是常數(P跟Q是須求的常數)*x的項] (2+p)=0... p=-2
and 1+p=Q Q=-2+1 Q=-1

2.
(-2)a-b+8=0...(1)
(-2)b-a+1=0...(2)
2[(-2)a-b+8]-[(-2)b-a+1]=0+0 <==2(1)-(2)
-4a-2b+16+2b+a-1=0
-3a+0b+15=0
3a=15
a=5...(3)
Put (3) into (1)
(-2)(5)-b+8=0
-10-b+8=0
-b=2
b=-2

7x² , 5x² , x² 係同類項...因為大家都有x²項
5x , 4x , x 係同類項...因為大家都有x項