中二恆等式習題

圖片參考：http://img832.imageshack.us/img832/1917/screenhunter03oct062308.gif


http://img832.imageshack.us/img832/1917/screenhunter03oct062308.gif

LHS=left hand side



證明下列各方程是否恆等式。
1.(x+5)(x-2)+x(x-3)=2(x-2)²+3x-5


LHS=(x+5)(x-2)+x(x-3)
=x²-2x+5x-10+x²-3x
=2x²-10
RHS=2(x-2)²+3x-5
=2(x²-4x+4)+3x-5
=2x²-8x+8+3x-5
=2x²-5x+3
LHSnot=RHS

Therefore(x+5)(x-2)+x(x-3)=2(x-2)²+3x-5 is not an identity.

2.(x+4/2)-(x+3/3)=1+(x/6)

LHS=(x+4/2)-(x+3/3)or((x+4)/2)-((x+3)/3)
=x-x+4/2-1 or (3x+12)/6-(2x+6)/6
=1 or (x+6)/6
RHS=1+(x/6)
=6/6+(x/6)
=(6+x)/6

LHS=RHS or not

Therefore(x+4/2)-(x+3/3)=1+(x/6) is not an identity or (x+4/2)-(x+3/3)=1+(x/6) is an identity.


求下列各恆等式中常數A及B的值。

1.Ax-3≡Bx-A

LHS=Ax-3

RHS=Bx-A

ThereforeAx-3≡Bx-A
-3=-A Bx=Ax
A=3 B=A
B=3



第二題唔識做~~



3.4x+A+1≡(B-1)x+4

LHS=4x+A+1

RHS=(B-1)x+4
=Bx-x+4

Therefore4x+A+1≡Bx-x+4
4x=Bx-x A+1=4
4=B-1 A=3
B=5