F. 2 Maths -Identities (最佳解答取20分)

1.) 3(x^2-1)+Ax+15
=3x^2-3+Ax+15
=3x^2+Ax+12

(B-x)(Cx-2)=BCx-2B-Cx^2+2x
=-Cx^2+(2+BC)x-2B

Hence,-2B=12
B=-6

-C=3
C=-3

2+BC=A
2+(-6)(-3)=A
A=20

So,A=20,B=-6,C=-3

2)If x^2+y^2=58 and xy=21, find the values of (x+y)^2 and (x-y)^2.
x^2+y^2=58
(x+y)^2-2xy=58
(x+y)^2=58+2xy
(x+y)^2=58+2(21)
(x+y)^2=58+42
(x+y)^2=100

x^2+y^2=58
(x-y)^2+2xy=58
(x-y)^2=58-2xy
(x-y)^2=58-2(21)
(x-y)^2=58-42
(x-y)^2=16

So,(x+y)^2=100 and (x-y)^2=16

1.) 3(x^2-1)+Ax+15
=3x^2-3+Ax+15
=3x^2+Ax+12

(B-x)(Cx-2)=BCx-2B-Cx^2+2x
=-Cx^2+(2+BC)x-2B

Hence,-2B=12
B=-6

-C=3
C=-3

2+BC=A
2+(-6)(-3)=A
A=20

So,A=20,B=-6,C=-3

2)If x^2+y^2=58 and xy=21, find the values of (x+y)^2 and (x-y)^2.
x^2+y^2=58
(x+y)^2-2xy=58
(x+y)^2=58+2xy
(x+y)^2=58+2(21)
(x+y)^2=58+42
(x+y)^2=100

x^2+y^2=58
(x-y)^2+2xy=58
(x-y)^2=58-2xy
(x-y)^2=58-2(21)
(x-y)^2=58-42
(x-y)^2=16

So,(x+y)^2=100 and (x-y)^2=16

1)3(x^2-1)+Ax+15=(B-x)(Cx-2)
sub X=0, -3+15=2B
B=6;
sub B=6,sub x=1,3(1-1)+A+15=(6-1)(C-2)
A-5C+25=0..........(1)
sub B=6,sub x=1,3(4-1)+2A+15=(6-2)(2C-2)
A-4C+16=0..............(2)
(2)-(1),C-9=0
C=9;
sub C=9 into (2),A=20.

2) (x+y)^2= x^2+y^2+2xy
=58+2(21)
=100;

(x-y)^2=x^2+y^2-2xy
=58-2(21)
=16;

1. 3(x^2-1)+Ax+15=(B-x)(Cx-2)
3x^2+Ax+12=-Cx^2+2x+BCx-2B
therefoere C=3..A=2..
henes,,BCx-2B=12
YOU FIND B YOURSELEVES!
2. XY=21
therefore X=3,Y=7
x^2+y^2
=3^2+7^2
=58

(x+y)^2
=(3+7)^2
=100
(x-y)^2
=(3-7)^2
=16