1.ax^2+xya+1/4y^2a 2.6x^4-8x^2y^2+y^4 3.4(3a+2x)^2-9(a-x)^2 4.(3m-x)^2-(m+3x)^2

1.ax^2+xya+1/4y^2a 2.6x^4-8x^2y^2+y^4 3.4(3a+2x)^2-9(a-x)^2 4.(3m-x)^2-(m+3x)^2

1. Take the formula a
Original formula = a × (x ^ 2 + XY + 1 / 4 y ^ 2) = a (x + 1 / 2 y) ^ 2
2. The original formula is 16x ^ 4-8x ^ 2Y ^ 2 + y ^ 4 = = >?
If so, it's (4x ^ 2-y ^ 2) ^ 2
3、【2（3a+2x）+3（a-x）】×【2（3a+2x）-3（a-x）】
=（9a+x）×（3a+5x）
4. (3m-x + m + 3x) × (3m-x-m-3x) = (4m + 2x) × (2m-4x)
=4（2m+x）*(m-2x)

1. (x + a) (x + 1) (x + 2) (x + 3) if the coefficient of the cubic term containing x is 17, then a is
2. If x + y = 4, x square + y square = 12, then (X-Y) square / XY=
3. If only x square + y square - 2x-4y + 5 = 0, then 1 / XY + 1 / (x + 1) (y + 1) + 1 / (x + 2) (y + 2) +. + 1 / (x + 2010) (y + 2010)
Especially questions 2 and 1

1. The cubic coefficient of X is a + 1 + 2 + 3, so a + 1 + 2 + 3 = 17, so a = 11 2. X + y = 4, the square of both sides is x square + y square + 2XY = 16, and x square + y square = 12, ∥ xy = 2, ∥ (X-Y) square / xy = (16-4 * 2) / 2 = 43. ∵ x square + y square - 2x-4y + 5 = 0, ∥ (x-1) square + (Y-2) square = 0, ∥

1) When factoring polynomial x ^ 2 + ax + B, Xiao Ming misread a and got (x-3) (x-4) after factoring; Xiao Li misread B and got (x-1) (X-7) after factoring. Find the original polynomial and the correct factoring result
2) It is known that the factorization result of the quadratic trinomial 2x ^ 2 + MX + n is (2x-1) (x + 0.25) to find the value of M and n

1) When factoring the polynomial x ^ 2 + ax + B, Xiao Ming misjudged a and got (x-3) (x-4) after factoring;
x^2+ax+b=x^2-7x+12
b=12
Xiao Li misunderstood B and got (x-1) (X-7) after decomposition
x^2+ax+b=x^2-8x+7
a=-8
Original polynomial x ^ 2-8x + 12 = (X-2) (X-6)
2) It is known that the factorization result of the quadratic trinomial 2x ^ 2 + MX + n is (2x-1) (x + 0.25) to find the value of M and n
2x^2+mx+n=（2x-1)(x+0.25)
2x^2+mx+n=2x^2+1/2x-x-0.25
2x^2+mx+n=2x^2-0.5x-0.25
m=-0.5 n=-0.25

(1-1 / 2 square) * (1-13 Square) * (1-1 / 4 square) * (1-1 / 5 square) * (1-1 / 6 square) * (1-1 / 7 square) * (1-1 / 8 square) * (1-1 / 9 square) * (1-1 / 10 square)

(a + b) (a-b) = the square of a minus the square of B

2010*2010-2009*2009+2008*2008···················+2*2-1*1
I don't understand

The original formula = (2010 square - 2009 Square) +. + (2 square - 1 square)
When you get to this step, use the square difference formula. It's in the first book of junior high school. After using this formula, it becomes
=【（2010+2009）（2010-2009）】+【（2008+2007）（2008-2007）】+.+【（2+1）（2-1）】
At this point, we found that the results of 2010-2009, 2008-2007, 2-1 were all 1
=（2010+2009）+（2008+2007）+.+（2+1）
At this point, it's very simple. It's 1 + 2 + 3 +. + 2010. Use the arithmetic summation formula (if you don't know, Baidu search, it's very simple)
=（1+2010）*2010/2=2021055

If the lengths of three sides of a triangle are a, B and C, and a ^ 2b-a ^ 2C + B ^ 2C-B ^ 3 = 0, then the triangle is
A. Isosceles B. right angles C. equilateral D. shape uncertainty
I know this problem should be factorized, but the shape of the triangle I made may be isosceles or equilateral. Isn't that D, but it's wrong and I can't figure it out~

The factorization result of polynomial 1-6b ^ 2 + 9b ^ 4 is

A group of passengers take a car. It is required that the number of passengers in each car is equal. At first, each car takes 22 people, but one person is left behind. If one car leaves empty, then all passengers can ride on the other cars equally. It is known that each car can hold up to 23 people. Question: how many cars are there at first? How many passengers are there

It's very simple. You can only take 23 people at most. It used to be 22 people per car. Now it's one less car. On the contrary, it's just average. The number of people per car is bound to increase, but it can't exceed 23 people. That is to say, it's just 23 people per car,
The equation is 22 * x + 1 = 23 * (x-1)
We can get x = 24, the number of people is 23 * 23 = 529

1-a＋2／a-1

Analysis:
A ≠ 1
(a+2)/(a-1)
=(a-1+3)/(a-1)
=1+ 3/(a-1)
If the value of fraction (a + 2) / (A-1) is an integer, then 3 / (A-1) must be an integer
That is, 3 can be divided by A-1
Then the possible values of such an integer A-1 are 3,1, - 1, - 3
So the corresponding values of a are: 4,2,0, - 2

Given 1a − 1b = 5, find the value of 2A + 3AB − 2BA − 2Ab − B

∵ 1a-1b = B − AAB = 5, ∵ B-A = 5ab, that is, A-B = - 5ab, then 2A + 3AB − 2BA − 2Ab − B = 2 (a − b) + 3AB (a − b) − 2Ab = − 10ab + 3AB − 5ab − 2Ab = − 7ab − 7ab = 1