✔ 最佳答案
1)首先,我明白你係點計
不過有一條係寫錯左
你係將佢expand哂之後再對位ga ma,right?
P(x-1)(x-2)+Q(x-2) (x+1)+R(x+1)(x-1)
=P(x^2-3x+2)+Q(x^2-x-2)+R(x^2-1)
=(P+Q+R)x^2-(3P+Q)x+(2P-2Q-R)
=4x^2-8x+6
SO....
P+Q+R=4..........(1)
3P+Q=8............(2)
2P-2Q-R=6........(3)
見唔見到同你個(2)有唔同??
你可以照做落去...
(1)+(2)+(3):
6P=18
P=3
sub P=3 into (2).
3(3)+Q=8
Q=-1
sub P=3,Q=-1 into (1)
3+(-1)+R=4
R=2
其實仲有另一個方法....
since P(x-1)(x-2)+Q(x-2)(x+1)+R(x+1)(x-1)≡4x^2-8x+6
你見到佢岩其實可以sub d數落去做,especially你見到有(x-2)(x-1)呢d,一sub個2/1落去,佢就變左做0,咁成個就冇左(因為任何數x0=0)
when x=-1,
P(-1-1)(-1-2)+Q(-1-2)(-1+1)+R(-1+1)(-1-1)=4(-1)^2-8(-1)+6
P(-2)(-3)=4+8+6
6P=18
P=3
when x=1,
P(1-1)(1-2)+Q(1-2)(1+1)+R(1+1)(1-1)=4(1)^2-8(1)+6
Q(-1)(2)=4-8+6
-2Q=2
Q=-1
when x=2,
P(2-1)(2-2)+Q(2-2)(2+1)+R(2+1)(2-1)=4(2)^2-8(2)+6
R(3)(1)=16-16+6
3R=6
R=2
2.都係一樣啦~
今次你expand係岩ge
咁其實都係照對位ge姐
(A+B+C)x^2+(2A-B-4C)x+(A-2B+4 C)=15x-12
so...
A+B+C=0.............(1)
2A-B-4C=15..........(2)
A-2B+4C=-12........(3)
(1)+(2):
3A-3C=15
A-C=5...........(4)
(1)X2:
2A+2B+2C=0...(5)
(5)+(3);
3A+6C=-12
A+2C=-4......(6)
(4)-(6):
-3C=9
C=-3
sub C=-3 into (4)
A-(-3)=5
A=2
sub A=2,C=-3 into (1)
2+B+(-3)=0
B=1
又或者用呢個方法.....
since A(x+1)^2+B(x+1)(x-2)+C(x-2)^2≡15x-12,
when x=-1,
A(-1+1)^2+B(-1+1)(-1-2)+C(-1-2)^2=15(-1)-12
C(-3)^2=-27
9C=-27
C=-3
when x=2,
A(2+1)^2+B(2+1)(2-2)+C(2-2)^2=15(2)-12
A(3)^2=18
9A=18
A=2
since A=2,C=-3,
when x=0,
2(0+1)^2+B(0+1)(0-2)+(-3)(0-2)^2=15(0)-12
2(1)^2+B(1)(-2)+(-3)(-2)^2=-12
2-2B-12=-12
-2B=-12+12-2
-2B=-2
B=1
2008-02-17 18:22:17 補充：
SO....1)P=3,Q=-1,R=22)A=2,B=1,C=-3
