Factorization AB + AC-B ^ 2-2bc-c ^ 2 2. Factorization 2 (a ^ 2-3bc) + a (4b-3c) 3. Known x + 2Y = 1 / 3, x-5y = - 1 / 2, Find the value of 6x ^ 2-18xy-60y ^ 2 4. Given 5x ^ 2-4xy + y ^ 2-2x + 1 = 0, find the value of (X-Y) ^ 2010 5. If the side length of a triangle is a, B, C are all integers, and a + BC + B + Ca = 8, find the perimeter of the triangle 6. Observe the following formula 3 ^ 2-1 ^ 2 = 8 = 8 * 1 5 ^ 2-3 ^ 2 = 16 = 8 * 2 7 ^ 2-5 ^ 2 = 24 = 8 * 3 9 ^ 2-7 ^ 2 = 32 = 8 * 4 (1) What law can you find from it? Try to express this Law in algebraic form (2) Can we write 1 + 2 + 3 + by using a series of formulas of reaction law in the first question +The formula of n (n is a positive integer)

Factorization AB + AC-B ^ 2-2bc-c ^ 2 2. Factorization 2 (a ^ 2-3bc) + a (4b-3c) 3. Known x + 2Y = 1 / 3, x-5y = - 1 / 2, Find the value of 6x ^ 2-18xy-60y ^ 2 4. Given 5x ^ 2-4xy + y ^ 2-2x + 1 = 0, find the value of (X-Y) ^ 2010 5. If the side length of a triangle is a, B, C are all integers, and a + BC + B + Ca = 8, find the perimeter of the triangle 6. Observe the following formula 3 ^ 2-1 ^ 2 = 8 = 8 * 1 5 ^ 2-3 ^ 2 = 16 = 8 * 2 7 ^ 2-5 ^ 2 = 24 = 8 * 3 9 ^ 2-7 ^ 2 = 32 = 8 * 4 (1) What law can you find from it? Try to express this Law in algebraic form (2) Can we write 1 + 2 + 3 + by using a series of formulas of reaction law in the first question +The formula of n (n is a positive integer)

1.=a(b+c)-(b+c)^2=(a-b-c)(b+c) 2.=2a(a+2b)-3c(a+2b)=(2a-3c)(a+2b)
3. = 6 (x-5y) (x + 2Y) = - 1 4. = (2x-y) ^ 2 + (x-1) ^ 2 = 0, so x = 1, y = 2x = 2, the evaluation = 1
5. A + BC + B + Ca = (a + b) (c + 1) = 8, a + B = 4, C + 1 = 2, perimeter = 5, or a + B = 2, C + 1 = 4. (shed) or C + 1 = 1 (shed) or C + 1 = 8 (shed)
6.(2k+1)^2-(2k-1)^2=8k
1+2+…… +n=[(3^2-1^2)+(5^2-3^2)+…… +((2n+1)^2-(2n-1)^2)]/8=[(2n+1)^2-1^2]/8=n(n+1)/2

-X & sup2; - 4Y & sup2; + 4xy 3ax & sup2; + 6AX + 3ay & sup2; 9 / 9 M & sup2; n & sup2; + 3 / 3 2Mn & sup2
The last point of the last question has not been typed out. The question is actually like this: 9 / M & sup2; n & sup2; + 3 / 2m n cubic + n quartic

-x²-4y²+4xy
=-(x²-4xy+4y²)
=-(x-2y)²
3ax²+6axy+3ay²
=3a(x²+2xy+y²)
=3a(x+y)²
9% M & sup2; n & sup2; + 3% 2Mn & sup2;
=Mn & sup2; (M + 6)

Factorization MX ^ 3-mx ^ 2-mx + M

=mx(x^2-1)-m(x^2-1)
=m(x-1)(x+1)(x-1)
=m(x+1)(x-1)^2

Factorization: given that x ^ 2 + MX + 5 has a factor X + 1, find the value of M

The other is x + 5, M = 6