If a = {x ｛ x ｛ 178; + 3x-10

If a = {x ｛ x ｛ 178; + 3x-10

x²+3x-10

A = {x | x2 + ax + 1 = 0}, B = {x | x2-3x + 2 = 0}, and a is included in B, the value range of a is

x^2-3x+2≤0
(x-1)(x-2)≤0
1≤x≤2
So B = {x | 1 ≤ x ≤ 2}
x^2+ax+1≤0
(1)a^2-4

If 3x2a + B + 5ya-2b-2 = 0 is a quadratic equation with respect to the letters X and y, then ab=______ ．

Because 3x2a + B + 5ya-2b-2 = 0 is a quadratic equation with respect to the letters X and y, then 2A + B = 1 and a-2b-2 = 1, the solution is a = 1, B = - 1, so AB = 1

Factorization of 2x (a-b) - (B-A)

Original formula = 2x (a-b) + (a-b)
=(a-b)(2x+1)

Factorization 2x (a-b) - 6 (B-A)

=2x（a-b）+6（a-b）=（2x+6）（a-b）

Factorization of 2x (a-b) - 4Y (B-A)

2x(a-b)-4y(b-a )
=2x(a-b)+4y(a-b )
=2(a-b)(x+2y)

Factorization (x + a) 3 + (x + b) 3 - (2x + A + b) 3

Several factorizations
（1）x³+x²+2xy+y²+y³
（2）x³+3x²y+3xy²+y³
（3）x³-3x²y+3xy²-y³
（4）x³+3x²+3x+2

(1)(x³+y³)+ (x²+2xy+y²)=(x+y)(x²-xy+y²)+(x+y)(x+y)=(x+y)(x²-xy+y²+x+y)
(2)(x³+y³)+3xy(x+y)=(x+y)(x²-xy+y²)+3xy(x+y)=(x+y)(x+y)(x+y)
(3)(x³-y³)-3xy(x-y)=(x-y)(x²+xy+y²)-3xy(x-y)=(x-y)(x-y)(x-y)
(4)(x³-1)+3x²+3x+3=(x-1)(x²+x+1)+3(x²+x+1)=(x+2)(x²+x+1)

To find these factorizations,
1.4A^-9B^+C^-4AC
2.(AX+BY)^+(BX-AY)^
3.X^(X-Y)+(Y-X)
4.(X-3)(X-1)+1
5.P(X-Y)-Q(Y-X)
6.(X+Y)^-(Y+X)
Note: ＾ means square

(a+b)^4-(a+b)^2
x^2(x-y)+y^2(x-y)
(7m+3n)^2-(5m+7n)^2

1、(a+b)^4-(a+b)^2=＜（a+b)^ 2+a+b)＞＜（a+b)^2-a-b＞=(a+b)(a+b+1)＜（a+b)^2-a-b＞2、x^2(x-y)+y^2(x-y)=(x-y)(x^2-y^2)=(x-y)(x+y)(x-y)=(x+y)(x-y)^23、(7m+3n)^2-(5m+7n)^2=(7m+3n+5m+7n)(7m+3n-5m-7n)=(12m...