恆等式"(3項)

1. L.H.S. = Cx^2 + Ax + (A - B)
R.H.S.
= 2x+6 + 3x-12
= 5x-6
Cx^2 + Ax + (A-B) = 5x-6
C = 0
A = 5
A-B = -6
5-B = -6
B = 11

2. L.H.S.
= 16x^2 - 24x + 9 + A
= 16x^2 - 24x + (A+9)
R.H.S.
= B(x+5)(2x-3) + Cx
= B(2x^2+10x-3x-15) + Cx
= B(2x^2+ 7x - 15) + Cx
= 2Bx^2 + 7Bx - 15B + Cx
= 2Bx^2 + (7B+C)x - 15B
16x^2 - 24x + (A+9) = 2Bx^2 + (7B+C)x - 15B
2B = 16 , B = 8
7B+C = -24
7(8)+C = -24
C = -80
A+9 = -15B
A+9 = -15(8)
A = -129
3. L.H.S.
= Ax(x-1) + B(x+1)(x-1) + C(x+1)x
= A(x^2-x) + B(x^2-1) + C(x^2+x)
= Ax^2 - Ax + Bx^2 - B + Cx^2 + Cx
= (A+B+C)x^2 + (C-A)x - B
(A+B+C)x^2 + (C-A)x - B = x^2+4
-B = 4
B = -4
A+B+C = 1
A+C-4 = 1
A+C = 5 ---(1)
C-A = 0
A = C ---(2)
Put (2) into (1)
C+C = 5
2C = 5
C = 5/2
A = C = 5/2

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Cx^2 + A(x+1)-B = 2(x+3)+3(x-4)

LHS=Cx^2 + A(x+1)-B
=Cx^2+Ax+(A-B)

RHS=2(x+3)+3(x-4)
=2x+6+3x-12
=5x-6

比較同類項,可得:
C=0
A=5

A-B=-6
5-B =-6
-B =-11
B=11

2010-04-24 14:37:41 補充：
(4x-3)^2 + A = B(x+5)(2x-3) + Cx
LHS=(4x-3)^2 + A
=16x^2-24x+(9+A)

RHS= B(x+5)(2x-3) + Cx
=B(2x^2+7x-15)+Cx
=2Bx^2+7Bx-15B+Cx

比較同類項,可得:
2B=16
B =8

C+7B=-24
C+56=-24
C =-80

9+A=-15B
9+A=-120
A =-129

2010-04-24 14:51:52 補充：
Ax(x-1) + B(x+1)(x-1) + C(x+1)x = x^2 + 4

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

A-C=0
A =C

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

so.A=2.5 B=-4 C=2.5

Question-1.

LHS
= Cx^2 + A(x+1) - B
= Cx^2 + Ax + A - B
= Cx^2 + Ax + (A - B)

RHS
= 2(x+3) + 3(x-4)
= 2x + 6 + 3x - 12
= 5x - 6

When comparing the coefficients of like terms,

C = 0

A = 5

A-B = -6
5-B = -6
B = 11

*********************************************************************

Question-2.

LHS
= (4x-3)^2 + A
= (4x)^2 - 2(4x)(3) + (3)^2 + A
= 16x^2 - 24x + 9 + A
= 16x^2 - 24x + (9+A)

RHS
= B(x+5)(2x-3) + Cx
= B(2x^2 - 3x + 10x - 15) +Cx
= (2B)x^2 + (7B)x - 15B + Cx
= (2B)x^2 + (7B+C)x - 15B

When comparing the coefficients of like terms,

2B = 16
B = 8

7B+C = -24
7(8)+C = -24
C = -80

9+A = -15B
9+A = -15(8)
9+A = -120
A = -129

*********************************************************************

Question-3.

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

RHS
= x^2 + 4

When comparing the coefficients of like terms,
-B = 4
B = -4

A-C = 0
A = C

A+B+C = 1
2A-4 = 1
2A = 5
A = 2.5

Since A = 2.5 and A = C
Therefore C = 2.5.