F.2 MATH 急~THX!

1
(x-3)^2-1≡(x-4)(x-2)
left=[(x-3)-1][(x-3)+1]
left=(x-4)(x-2)
left=right
所以(x-3)^2-1≡(x-4)(x-2)是恆等式。

2
(x+7)(2x-5)≡Ax^2+Bx+C
left=2x^2-5x+14x-35
left=2x^2+9x-35
A=2 B=9 C=-35

a.
(x-4)(x+4)
=x^2-16

b.
(-x+3)(3+x)
=(3-x)(3+x)
=9-x^2

c.
(4x-1)(1+4x)
=(4x-1)(4x+1)
=16x^2-1

d.
(x+2)^2
=x^2+4x+4

e.
(-3x-2)(3x-2)
=-9x^2+6x-6x+4
=-9x^2+4

f.
(3-h)^2
=9-6h+h^2

g.
-2(3x-4y)(-4y-3x)
=-2(-12xy-9x^2+16y^2+12xy)
=-2(16y^2-9x^2)
=-32y^2+18x^2

h.
(ab-cd)(ab+cd)
=a^2b^2-c^2d^2

i.
-2(ab+c)^2
=-2(a^2b^2+2abc+c^2)
=-2a^2b^2-4abc-2c^2

a.
6(a+1)^2+4(a+1)
=[2(a+1)][3(a+1)+2]
=2(a+1)(3a+3+2)
=2(a+1)(3a+5)

b.
4c^2-6dx+3cd-8cx
=c(4+3d)-2x(3d+4)
=(3d+4)(c-2x)

c.
4x^2-64y^2
=4(x^2-16y^2)
=4(x-4y)(x+4y)

d.
36-(x-y)^2
=[6-(x-y)][6+(x-y)]
=(-x+y+6)(x-y+6)

e.
(a+b)^2-(a-b)^2
=[(a+b)-(a-b)][(a+b)+(a-b)]
=2b(2a)
=4ab

f.
49-64x^2
=(7-8x)(7+8x)

g.
x^2+26x+169
=(x+13)^2

h.
x^2-18x+81
=(x-9)^2

i.
2x-1-x^2
=(x-1)^2

j.
16x^2+16x+4
=4(4x^2+4x+1)
=4(x+1)^2

1.證明(x-3)^2-1≡(x-4)(x-2)是恆等式。
(x^2-6x+9)-1=x^2-6x+8
x^2-6x+8=x^2-6x+8
so,(x-3)^2-1≡(x-4)(x-2)是恆等式。
2.(x+7)(2x-5)≡Ax^2+Bx+C，求A,B和C的値。
(2x^2-5x+14x-35)=Ax^2+Bx+C
2x^2+9x-35=Ax^2+Bx+C
so,A=2,B=9,C= -35

(x-4)(x+4)
=x^2-16
(-x+3)(3+x)
=9-x^2
(4x-1)(1+4x)
=16x^2-1
(x+2)^2
=x^2+4x+4
(-3x-2)(3x-2)
= -9x^2+4
(3-h)^2
=9-6h+h^2
-2(3x-4y)(-4y-3x)
= -2(-9x^2+16y)
= 18x^2-32y
(ab-cd)(ab+cd)
=a^2 b^2 - c^2 d^2
-2(ab+c)^2
= -2(a^2 b^2 +2abc+c^2)
= -2a^2b^2-4abc-c^2

2012-08-13 10:41:41 補充：
6(a+1)^2+4(a+1)
=2(a+1)(3a+3+4)
=2(a+1)(3a+7)
4c^2-6dx+3cd-8c
x=4c^2-8cx-6dx+3cd
=4c(c-2x)-3d(2x-c)
=(c-2x)(4c+3d)

2012-08-13 10:41:52 補充：
4x^2-64y^2
=4(x^2-16y^2)
=4(x+4y)(x-4y)
36-(x-y)^2
=(6+x-y)(6-x+y)
(a+b)^2-(a-b)^2
=(a+b+a-b)(a+b-a+b)
=2ax2b
=4ab
49-64x^2
=(7+8x)(7-8x)
x^2+26x+169
=(x+13)^2
x^2-18x+81
=(x-9)^2
2x-1-x^2
=-x^2+2x-1
=-(x^2-2x+1)
=-(x-1)^2
=(x-1)(1-x)
16x^2+16x+4
=4(4x^2+4x+1)
=4(2x+1)(2x-1)

1.證明(x-3)^2-1≡(x-4)(x-2)是恆等式。
L.H.S.=(x-3)^2-1=(x-3+1)(x-3-1)=(x-2)(x-4)=R.H.S.
2.(x+7)(2x-5)≡Ax^2+Bx+C，求A,B和C的値。
(x+7)(2x-5)=2x^2-5x+14x-35=2x^2+9x-35
Ax^2=2x^2-->A=2
Bx=9x-->B=9
C=-35
展開下列各式
a. (x-4)(x+4)=x^2-16
b. (-x+3)(3+x)=(3-x)(3+x)=9-x^2
c. (4x-1)(1+4x)=-(1-4x)(1+4x)=-(1-16x^2)=16x^2-1
d. (x+2)^2=x^2+4x+4
e. (-3x-2)(3x-2)=-(3x+2)(3x-2)=-(9x^2-4)=4-9x^2
f. (3-h)^2=9-6h+h^2=h^2-6h+9
g. -2(3x-4y)(-4y-3x)=2(3x-4y)(3x+4y)=2(9x^2-16y^2)=18x^2-32y^2
h. (ab-cd)(ab+cd)=a^2b^2-c^2d^2
i. -2(ab+c)^2=-2(a^2b^2+2abc+c^2)=-2a^2b^2-4abc-2c^2
因式分解下列各式
a. 6(a+1)^2+4(a+1)=2(a+1)(3a+3+4)=2(a+1)(3a+7)
b. 4c^2-6dx+3cd-8cx=4c^2-8cx-6dx+3cd=4c(c-2x)-3d(2x-c)=(c-2x)(4c+3d)
c. 4x^2-64y^2=4(x^2-16y^2)=4(x+4y)(x-4y)
d. 36-(x-y)^2=(6+x-y)(6-x+y)
e. (a+b)^2-(a-b)^2=(a+b+a-b)(a+b-a+b)=2ax2b=4ab
f. 49-64x^2=(7+8x)(7-8x)
g. x^2+26x+169=(x+13)^2
h. x^2-18x+81=(x-9)^2
i. 2x-1-x^2=-x^2+2x-1=-(x^2-2x+1)=-(x-1)^2=(x-1)(1-x)
j. 16x^2+16x+4=4(4x^2+4x+1)=4(2x+1)(2x-1)