A few factoring math problems! Ask for help!

4a(9a^2-1)

4a(3a+1)(3a-1)

2.x + quarter y

y/2)^2-x^2

y/2+x)(y/2-x)

4(a-b)+3(a+b)][4(a-b)-3(a+b)](7a-b)(a-7b)

to the 4th power - to the 4th power of y.

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

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

5.Factoring first, then evaluation:

x+y）²-x-y）²

x+y+x-y）（x+y-x+y）

2x*2y4xy

where x=1 y=2

4xy=46（x+p）²-x+q）²

x+p+x+q)(x+p-x-q)

2x+p+q)(p-q)

7.Four-ninths m

2m/3+

1. Original formula. 4a(9a²-1)

4a(3a+1)(3a-1)

2. Original formula. -[x²-(1/2y)²]

x+y/2)(x-y/2)

3. Original formula. [4(a-b)]²3(a+b)]²4a-4b+3a+3b][4a-4b-3a-3b](7a-b)(a-7b)

4. Original formula. （x²)²y²)²

x²+y²)(x²-y²)

x²+y²)(x+y)(x-y)

5. Original formula. （x+y）²-x-y）²

x+y+x-y)(x+y-x+y)

2x)(2y)4xy

6. Original formula. （x+p）²-x+q）²

x+p)+(x+q)][x+p)-（x+q)](2x+p+q)(p-q)

7. Original formula. Four-ninths m

2/3m)²

2/3m+

The first = 4a(9a2-1).Everything else is a problem, with the specific formula a 2-b 2=(a b)(a-b).

I think you made a mistake on the topic.

I guess I'm looking for the value of a2+1 a2.

a2+3a+1=0, obviously a is not 0, both sides are divided by aa+1 a=-3, both sides are squared, a2+2+1 a2=9, a2+1 a2=7

a 2 = 1 a 2 then a = 1 or a = -1

However, substituting a 2 + 3 a + 1 = 0 does not hold.

So see if you get the title wrong!

1； 5（x-y）（x-y+2)

2； （a+b)^2+12(a+b)+36=(a+b+6)^23；m 2+5n-mn+5m=m(m+n) should be problematic??

4； 4y(x+2)-2z(x+2)=2(2y-z)(x+2)

It's very simple.

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

1） 2/(2a+3)+3/(3-2a)+（2a+15）/(4a²-9）

4a-6）/（4a²-9）-（6a+9）/(4a²-9）+（2a+15）/(4a²-9）

4a-6-6a-9+2a+15）/(4a²-9）

2） 1/（x+3）+6/（x²-9）-（x-1）/（6-2x）

2（x-3）/2（x+3）（x-3)+6*2/2（x+3）（x-3)+（x-1）（x+3）/2（x+3)(x-3)

2x-6）+12+（x-1）（x+3）】/2（x+3)(x-3)

x²+2x-3+2x-6+12）/2（x+3)(x-3)

x²+4x+3）/2（x+3)(x-3)

x+3）（x+1)/2(x+3)(x-3)

x+1)/2(x-3)

3) (x-1/x+1 +x 2+2x-1/1) divided by x-1/2 x 2 Due to typing reasons, I changed the next question without authorization, and the meaning of the question remains unchanged.

x-1 of x+1 + x 2+2x-1/1) multiplied by x-1/2

x+1)/（x-1)+1/（x²+2x-1）】×x-1)/x²

x+1)(x-1)/（x-1)x²+（x-1）/（x²+2x-1）x²

x+1)/x²+（x-1）/（x²+2x-1）x²

x+1）（x²+2x-1）/（x²+2x-1）x²+（x-1)/（x²+2x-1）x²

x³+2x²-x+x²+2x-1+x-1）/（x²+2x-1）x²

x³+3x²+2x-2)/（x²+2x-1）x²

1) 2 (2a+3)+3 (3-2a)+(2a+15) (4a, 2-9) Is this the case with the first question? I really can't tell which is the denominator, and there are no parentheses.

2/(2a+3)+3/（3-2a）+（2a+15）/（4a^2-9）

2a+15）/（9-4a^2)+(2a+15）/（4a^2-9）

2)1/(x+3)+6/(x^2-9)-(x-1)/(6-2x)

x-3)/(x^2-9)+6/(x^2-9)-(x-1)/(6-2x)

1/(x-3)-(x-1)/(6-2x)

x+1)/(2x-6)

3)[(x+1)/(x-1)+1/(x^2+2x-1)]/(x^2)/(x-1)

x+1)/(x-1)+1/(x^2+2x-1)]*x-1)/(x^2)

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

x^3+3x^2+2x-2)/(x^4+2x^3-x^2)

4)[1+1/(x-1)]/x/(x^2-1)

1+1/(x-1)]*x^2-1)/x

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

x+15)[(2-x)/(x-1)]/[x+1-3/(x-1)

2-x)/(x-1)]/[(x^2-4)/(x-1)]

2-x)/(x-1)]*x-1)/(x^2-4)]

1/(x+2)

6)[(3x+3)/(x^2+x-2)]/[x-2+3/(x+2)-1(1-x)]

3x+3)/(x^2+x-2)]/[(x^3-x^2+3)/(x+2)(x-1)]

3x+3)/(x^3-x^2+3)