A factorization math problem! Urgent!!!!

The first question takes the minimum value when a=2 and b=1, the second question = 3 to the 16th power, the third question has a side length of 5, and the fourth question = -1, I am a sky walker

1.x 2-2(m-3)x+25 is a perfectly square method, which can be obtained according to the formula of the perfect square Li Wei Finch.

m-3 = 5，m-3 = 5

So, m=8, and the mountain bond m=-2

2, 25x（

5*5x ( 5*2y(

5( (5x-2y)

3. 1/2a^3b+a²b²+1/2ab^31/2ab (a^2+2ab+b^2)

1/2ab (a+b)^2

Put a+b=2, ab=2 generations which early into the above formula, get.

1/2ab (a+b)^2

The original formula = (x-y-1)(x+3y-4) can be obtained by multiplying the double cross

Question 1: -8a 0

a original = 2a (-2a).

Question 2: A 0, B 0

The original numerator denominator is divided by (ab) to obtain:

2/(√a-√b)

Then multiply the numerator and denominator by ( a + b) to get :

2(√a+√b)/(a-b)

Root number under -8a3 0

a3 is less than or equal to 0

a is less than or equal to 0

Paralysis Lao Tzu actually won't.

①x^4-12x+323

x^4+324-12x-1

x^4+36x^2+324)-(36x^2+12x+1)

x^2+18)^2-(6x+1)^2

x^2+18+6x+1)(x^2+18-6x-1)

x^2+6x+19)(x^2-6x+17)

x^4+7x^3+14x^2+7x+1

x^4+7x^3+12x^2 + 2x^2+7x + 1

x^2+3x)(x^2+4x) +x^2+3x + x^2+4x + 1

x^2+3x)(x^2+4x+1)+(x^2+4x+1)

x^2+3x+1)*(x^2+4x+1)

x 2-3x-3)(x 2+3x+4)-8 cannot be decomposed within the range of rational number coefficients, according to the factor theorem.

x^2y^2+1-x^2-y^2+4xy

x^2y^2+2xy+1-x^2-y^2+2xy

xy+1)^2-(x-y)^2

xy+1+x-y)(xy+1-x+y)

x^4+x^2+2ax+1-a^2

x^4+2x^2+1-x^2+2ax-a^2

x^2+1)^2-(x-a)^2

x^2+1+x-a)(x^2+1-x+a)

x+y)^4+x^4+y^4

x^2+2xy+ y^2)^2+x^4+y^4

x^4+y^4+4x^2y^2+2x^2y^2+4x^3y+4xy^3+ x^4+y^4

2[x^4+y^4+3x^2y^2+2xy(x^2+ y^2))]

2[x^4+y^4+2x^2y^2+2xy(x^2+ y^2)+ x^2y^2]

2[(x^2+y^2)^2+2xy(x^2+ y^2)+ xy)^2]

2(x^2+ y^2+xy)^2

a^3b-ab^3+a^2+b^2+1

a^3b-a^2b^2+a^2b^2+ab-ab-ab^3+a^2+b^2+1

a^3b-a^2b^2+ab+a^2b^2-ab^3+b^2+a^2-ab+1

ab(a^2-ab+1)+b^2(a^2-ab+1)+1(a^2-ab+1)

ab+b^2+1)(a^2-ab-1)