(1) (-8)+10+2+(-1) =2+2+(-1) =4+(-1) =3 (2) 5+(-6)+3+9+(-4)+

(3m-4n)^2+2(3m-4n)(3m+4n)-3(3m+4n)^2
=9m^2-24mn+16n^3+18m^2-32n^2-27m^2-72mn-48n^2
=-96mn-64n^2
=-96*(-2)*1/4-64*(1/4)^2
=48-4
=44

（5.2-3X÷2+2）÷3=0.5
5.2-3X÷2+2=0.5×3=1.5
3x÷2=5.2+2-1.5
3x÷2=5.7
x=5.7×2÷3
x=3.8

1.（x^2-3)^2+2(x^2-3)(x-3)+(x-3)^2
=(x^2-3+x-3)^2
=(x^2+x-6)^2
=(x+3)^2(x-2)^2
2.(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x)+22(x^2+8x)+105+15
=(x^2+8x)+22(x^2+8x)+120
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
3.x^2+y^2+4x+2y+3——题目是不是写错了呀
x^2-y^2+4x+2y+3
=(x^2+4x+4)-(y^2-2y+1)
=(x+2)^2-(y-1)^2
=(x+2+y-1)(x+2-y+1)
=(x+y+1)(x-y+3)
4.(a^2+a+1)^2+2(a^2+a+1)(a^2+a-1)+(a^2+a-1)^2
=(a^2+a+1+a^2+a-1)^2
=(2a^2+2a)^2
=[2a(a+1)]^2
=4a^2(a+1)^2
5.(x^2+4)^2-16x^2
=(x^2+4+4x)(x^2+4-4x)
=(x+2)^2(x-2)^2
6.(a+b+c)^2-(a-b-c)^2
=(a+b+c+a-b-c)(a+b+c-a+b+c)
=2a(2b+2c)
=4a(b+c)
7.25(x-y)^2-16(x+y)^2
=[5(x-y)+4(x+y)][5(x-y)-4(x+y)]
=(9x-y)(x-9y)
8.x^4-81x^2y^2
=x^2(x^2-9y^2)
=x^2(x+3y)(x-3y)
9.x^2(x-y)+y^2(y-x)
=(x^2-y^2)(x-y)
=(x+y)(x-y)^2
10.(a-b)(x-y)+(b-a)(x+y)
=(b-a)(-x+y+x+y)
=2y(b-a)
(乘法)
11.2(2m-1)^2-3(m+1)(m-1)-2(m-1)^2
=8m^2-8m+2-3m^2+3-2m^2+4m-2
=3m^2-4m+3
12.(2x+y+3t)(2x-y-3t)
=(2x)^2-(y+3t)^2
=4x^2-y^2-6ty-9t^2
13.(3a+1)^2(3a-1)^2
=[(3a+1)(3a-1)]^2
=(9a^2-1)^2
=81a^4-18a^2+1
14.(2x-y)(2x+y)-(3x-y)(4x-y)
=4x^2-y^2-(12x^2-7xy+y^2)
=-8x^2+7xy-2y^2
15.(-4x+2y)(-2y-4x)-(2x+3y)(8x-9y)
=(-4x)^2-(2y)^2-(16x^2+6xy-27y^2)
=16x^2-4y^2-16x^2-6xy+27y^2
=23y^2-6xy
16.(2a+b-c)^2
=(2a)^2+2·2a(b-c)+(b-c)^2
=4a^2+4ab-4ac+b^2-2bc+c^2
=4a^2+b^2+c^2+4ab-4ac-2bc

f(x)=x^2+2 (x≤2)
=2x (x>2)
1.x=-4,属于第一个关系式,所以f(-4)=(-4)^2+2=18
2.f(x0)=8,设x0≤2,则f(x0)=(x0)^2+2=8,
解出x0=-根号6或是根号6,因为假设x0≤2
所以x0=-根号6
设x0>2,则f(x0)=2x0=8,解出x0=4,符合假设
所以x0=4
最后可知x0=-根号6或x0=4

a^2+2ab+b^2-2ac+c^2-2bc
=(a^2+2ab+b^2)-2ac-2bc+c^2
=(a+b)^2-2(a+b)*c+c^2
=(a+b-c)^2

(1)(a+6b-4c)(a-2b+2c)
(2)原式=9x²-x²y²+y²-1+8xy
=9x²+6xy+y²-(x²y²-2xy+1)
=(3x+y)²-(xy-1)²
=(3x+y+xy-1)(3x+y-xy+1)
(3)(a-3b)(a-3ab+3b²)

3
4
6
8

c-a*(b/2a)^2
=c-a*b²/(4a²)
=c-b²/(4a)
=4ac/(4a)-b²/(4a)
=(4ac-b²)/(4a)

A={X|x^2+4x=0}={-4,0} , A交B=B,说明B是A的子集 B是二次方程 若无解，是空集，符合题意 判别式=4[(a+1)²-(a²-1)]0 a-1 判别式=0 则a=-1 方程是x²=0,x=0 符合B是A的子集 判别式大于0 a-1 此时有两个解 则A和B必须是同一个方程 所以2(a+1)=4 a²-1=0 a=1 综上 a≤-1，a=1

2(a^2+b^2)-|(a^2+2ab+b^2)-[a^2-(2ab-b^2]|
=2a^2+2b^2-|a^2+2ab+b^2-a^2+2ab-b^2|
=2a^2+2b^2-4|ab|
=2(|a|-|b|)^2
楼上的忘记了带绝对值的,不知对否?

2x*5x^3 =10x^4
3x^3*(-1/3x) =-x^4
3x*(-2x^2) =-6x^3
a*(x-y) =ax-ay
x^2*(1-x-x^2) =x^2-x^3-x^4
(a^2+2ab+3a^2b)=a^2+2ab+3a^2

=（a+b)^2-2（a+b)+1=（a+b-1)^2

x(x+4)(x-1/2)=0 解得x=0、-4或1/2,因x∈z,
所以 B={x l x=0或-4}
因A∩B=A,
所以A=空集或{0}或{-4}或{-4,0}
1、当A=空集时,方程x^2+2(a+1)+a^2-1=0无解
必须： （2a+2)^2-4(a^2-1)

(b x + a y) (a x + b y) (2 + (b x + a y) (a x + b y))
=2 a b x^2 + a^2 b^2 x^4 + 2 a^2 x y + 2 b^2 x y + 2 a^3 b x^3 y + 2 a b^3 x^3 y + 2 a b y^2 + a^4 x^2 y^2 + 4 a^2 b^2 x^2 y^2 + b^4 x^2 y^2 + 2 a^3 b x y^3 + 2 a b^3 x y^3 + a^2 b^2 y^4

2(a-2b)^2-7(a-2b)^3+3(2b-a)^2+2(2b-a)^3
=2(a-2b)^2-7(a-2b)^3+3(a-2b)^2-2(a-2b)^3
=5(a-2b)^2-9(a-2b)^3

原式=[2cos^3 θ+sin^2 θ+cosθ-3]/[2+2cos^2 θ+cosθ]
所以,f(π/3)=[2*(1/2)^3+(√3/2)^2+(1/2)-3]/[2+2*(1/2)^2+(1/2)]
=[2*(1/8)+(3/4)+(1/2)-3]/[2+2*(1/4)+(1/2)]
=(-3/2)/3
=-1/2

4

（1）（x+y-3)(x+y+3)
=[(x+y)-3][(x+y)+3]
=(x+y)²-9
=x²+2xy+y²-9
(2)(x-1)(x+1)(x^2-1)
=(x²-1)²
=x的4次方-2x²+1
(3)(x+2y)^2+2(x+2y)+1
=(x+2y+1)²

1.=2ab(2ab-3)
2.=3(x^2-1)=3(x-1)(x+1)
3.=x(x^2-1)=x(x-1)(x+1)
4.=a(a^2+4a+4)=a(a+2)^2
5.=a(a+x+1)
6.=-4x(x^2+4x+6)
7.=(3a)^4-(2b)^4=[(3a)^2+(2b)^2][(3a)^2(2b)^2]=[(3a)^2+(2b)^2][(3a-2b)(3a+2b)]
8.=(x+y+z)(x+y-z)
9.-0.01(100-x)
10.(x^2+1)(x^2-1)=(x^2+1)(x-1)(x+1)

先观察,这道题是利用完全平方公式解答的,所以你要一步一步的回答就行了a^2+2a(b+c)+(b+c)^2
=a^2+a(b+c)+a(b+c)+(b+c)^2
=a(a+b+c)+(b+c)(a+b+c)
=(a+b+c)^2

6+4√3-2√6
不知道对不对,嘿嘿

1/(cosx)^2+2/(sinx)^2
=[1/(cosx)^2+2/(sinx)^2]*[(sinx)^2+(cosx)^2] (反正(sinx)^2+(cosx)^2=1,乘了也没事)
=(sinx/cosx)^2+1+2+2(cos/sinx)^2
第一项,和最后一项用均值不等式就可以了
>=3+2√2

1.(a-4)^2+(b-4)^2+2(ab-8)
=a^2-8a+16+b^2-8b+16+2ab-16
=a^2+2ab+b^2-8a-8b+16
=(a+b)^2-8(a+b)+16
=(a+b-4)^2
2.ac^2(2b+3c)^2-ab^2(3c+2b)^2
=a[c(2b+3c)+b(3c+2b)][c(2b+3c)-b(3c+2b)] 提取工因式+平方差公式
=a（2b^2+5bc+3c^2）(3c^2-bc-2b^2)
=a(2b+c)(b+3c)(c-b)(3c+2b)

原式
= （√3 - √2） + （2 + 2√2）
= √3 + 2 + √2

A={X|x^2+4x=0}={-4,0} ,
A交B=B,说明B是A的子集
B是二次方程
若无解,是空集,符合题意
判别式=4[(a+1)²-(a²-1)]0
a-1
判别式=0
则a=-1
方程是x²=0,x=0
符合B是A的子集
判别式大于0
a-1
此时有两个解
则A和B必须是同一个方程
所以2(a+1)=4
a²-1=0
a=1
综上
a≤-1,a=1

=√3+1-2+1
=√3-1+1
=√3