f.2 math 1條 (identity)

Ax(x-1)+B(x+1)(x-1)+ C(x+1)x=x^2+4
Ax^2-Ax+Bx^2-B+Cx^2+Cx=x^2+4
(A+B+C)x^2+(C-A)x-B=x^2+4
B=-4
C-A=0---->A=C
2A-4=1
A=C=5/2

Ax(x-1)+B(x+1)(x-1)+C(x+1)x=x^2+4
Ax^2-Ax+B(x^2-1)+Cx^2+Cx=x^2+4
Ax^2-Ax+Bx^2-B+Cx^2+Cx=x^2+4
(A+B+C)x^2+(C-A)x-B=x^2+4

答 : B=-4 ,

A+B+C=1
A+(-4)+C=1
A+C=5----------(1)

C-A=0
-A+C=0-------------(2)

(1)-(2) , 2A=5
A=5/2

(1) , 5/2+C=5
C=5/2

Ax(x-1)+B(x+1)(x-1)+C(x+1)x=x^2+4
[(Ax^2)-Ax]+[(Bx+B)(x-1)]+[(Cx+C)x]=(x^2)+4
[(Ax^2)-Ax]+[(Bx^2)-Bx+Bx-B]+[(Cx^2)+Cx]=(x^2)+4
[(Ax^2)-Ax]+[(Bx^2)-B]+[(Cx^2)+Cx]=(x^2)+4
[(Ax^2)+(Bx^2)+(Cx^2)]-[Ax-Cx]-B=(x^2)+4

∴-B=4
B=-4

A-C= 0.........(1)
A+B+C=1
A-4+C=1
A+C=5............(2)
(2)-(1)
2C=5
C=5/2
Sub C=5/2 into (1)
A-5/2=0
A=5/2

∴A=5/2
B=-4
C=5/2