Ball math problems, hurry! Tomorrow is handed in, the first year of the new year, decomposition fact

1) Multiply the square of (x+y) by (x+y-1).

2) Multiply the square of (x-y) by (x-y-1).

3) Multiply the square of (x-y) by (x-y+).

4) (x-y) multiplied by (2-3x+3y).

5) 2ab(a+b)(2a+2b-3a)=2ab(a+b)(2b-a)

1. Let x+y a, that is, the square of a (a 2) minus the cube of a (a 3), the common factor is the square of a (a 2), extract the common factor a 2 to obtain (1-a), and then substitute a=x+y into (x+y) 2(1-x-y); This method is called the commutation method, and the same method can be used to decompose the factor.

2、（x-y)^2(1-x+y)

3、（x-y)^2(1+x+y)

4、(x-y)(2+3x+3y)

5. After extracting the common factor 2AB first, 2AB*2(a+b) 2-2AB*3A(A+B) is obtained, and then extracting the common factor (A+B) to obtain 2AB(A+B)*2(A+B)-2AB(A+B)*3A, the two formulas are combined into 2AB(A+B)(2(A+B)-3A), which is simplified to 2AB(A+B)(2B-A).

1.(x-4)(x+3)

What is A-1 6A-6 1? Orange acres.

x+5)(x+5)

x-3)(x-8)

m+9)(m-2)

p-9)(p+4)

t-4)(t+2)

x-2)(x+6)

m-3)(m-16)

7 in n -7 n -60 should be 17, right?

2.(a+b)(a+b)=2*2=4

Get: x=y

4.(16n*n+40n+16) 8=2n*n+5n+2 is the number of whole families. So the original proposition of Eneda Mori is proven, 5∵(a-b)(a-b)+(b-c)(b-c)+(c-a)(c-a)=0

a=b=c

2. d.-42

Calculation problem = m (a-2) - m (a-2).

m（a-2）(m-1)

a（x-y）-b（x-y）²

x-y)[a-b(x-y)]

x-y)(a-bx+by)

7.If x+y=3, xy=-4. Find the value of 2x y+2xy.

Calculated with an easy method).

2x²y+2xy²

2xy(x+y)

410.Knowing x +x-1=0, find the value of x +2x +2009 x +x-1=0

x²+x=1

x³+2x²+2009

x³+x²+x²+2009

x(x²+x)+x²+2009

x+x²+2009

①d②d

c a5, original = m (a-2)-m (a-2).

m（a-2)（m-1)

6. Original formula = a(x-y)-b(x-y).

x-y)（a-bx+by）

7. Original formula = 2xy(x+y).

8. Original = 9, original = 410, x + x-1 = 0

x²+x=1

So the original formula =x +x +x +2009

x(x +x)+x +2009 [Replace it with 1 when you see x +x] = x+x +2009

1.It is known that one factor of the polynomial 2x -y -xy-x-2y-1 is (x-y-1) and the other factor is ax+by+cFind the value of (a+b)·c.

Solution: Derived from the question:

x-y-1)×(ax+by+c)=ax²-axy-ax+bxy-by²-by+cx-cy-c

ax²-by²+(b-a)xy+(c-a)x+(-b-c)y-c

2x²-y²-xy-x-2y-1

So: a=2, b=1, c=1

a+b)·c=(2+1)×1=3

2.Knowing x+y=6,xy=-6, find the value of 4(x-y) -x-y) 4+760.

Solution: Because (x-y) = (x+y) -4xy=6 -4 (-6)=60

So (x-y) 4=[(x-y)]=3600

So 4(x-y) -x-y) 4+760

3.It is known that a, b, and c are the three sides of the triangle, and they satisfy a + b + c -ab-bc-ac = 0It is to determine what kind of triangle this triangle is, and to explain why.

Solution: This triangle is an equilateral triangle.

Reason: Because a +b +c -ab-bc-ac=0, both sides are multiplied by 2.

2a²+2b²+2c²-2ab-2bc-2ac=0

a²-2ab+b²+a²-2ac+c²+b²-2bc+c²=0

a-b)²+a-c)²+b-c)²=0

Because (a-b) 0, (a-c) 0, (b-c) 0

then (a-b) = 0, (a-c) = 0, (b-c) = 0

So a-b=0, a-c=0, b-c=0

So: a=b, a=c, b=c

i.e.: a=b=c

So this triangle is an equilateral triangle.

A=2 B=1 C=1

2. (x+y）²=36

x-y）² = ( x+y）²-4xy=36 - 4(-6) =60

x-y）^4= ( x-y）² x-y）² =3600

4（x-y）²-x-y）^4+760=4*60-3600+760=-2600

3. a²+b²+c²-ab-bc-ac=0 2a²+2b²+2c²-2ab-2bc-2ac=0

a-b）²+b-c）²+a-c）²=0

So a=b=c The triangle is equilateral.

1.Pending coefficient method (x-y-1) (ax+by+c) = 2x -y -xy-x-1

Get ax +bxy+cx-axy-by +cy-ax-by-c=2x -y -xy-x-1

Comparing the coefficients of the quadratic term and the constant term, a=2, b=1, c=1 (a+b)c=3 are obtained

2. Find (x-y) first, x-y) =x +y -2xy=(x+y) -4xy=60

Substituting 4(x-y) -x-y) 4+760 results in -2600 ((x-y) 4 is its fourth power, right? ）

3 equilateral, 2(a +b +c -ab-bc-ac) = (a-b) +b-c) +c-a) = 0

Obviously (a-b) 0, (a-c) 0, (c-b) 0

So there can only be (a-b) =(b-c) =(a-c) =0

a=b=c is an equilateral triangle.

Give it points Write for a long time, 3q

1, the generally known polynomial = (x-y-1)*(ax+by+c), now the question is to find the value of a, b, c, observe the equation I wrote, the previous factor and the latter factor are multiplied where abc will appear in x*ax, (-y)*by, (-1)*c, and these are equal to the coefficients in front of the quadratic terms of the known polynomial, such as: 2x, -y, and -1, so the values of a, b, and c are 2, 1, 1 respectively.

2. According to the known conditions, first find (x-y) =(x+y) -4xy=36-4*(-6)=60, and calculate it by substitution: 60-60*60+760=2780.

3, 2a +2b +2c -2ab-2bc-2ac = 0After merging(a -2ab+b )+a -2ac+c )+b -2bc+c )=0, then (a-b) +a-c) +b-c) =0 is obtained, so that a=b=c, so the triangle is equilateral.

Since my level is limited, please point out any mistakes!

1. 2x²-y²-xy-x-2y-1=(x-y-1)（2x+y+1） a=2,b=1,c=1,(a+b)·c=3

2.Because x+y=6,xy=-6, (x-y) =60,4(x-y) -x-y) 4+760=-2600

3.Because a + b + c -ab-bc-ac = 0, (a-b) (b-c) (c-a) = 0, the triangle is an isosceles triangle.

-x square y + 8xy - 15y = -x (x-3) (x-5) x to the 2n power - 5x to the n power - 66 = (x n + 6) (x n - 11) 3 to the 2m power + 3 m + 2 to the power + 81 quarters = (3 m + a m + 2 power + 3a square x - 2a m power y-6xy = (a 2-2y) (a m + 3a 2 x).

x to the 2n power + 6x to the n power - y squared + 9 = (x n + y + 3) (x n - y + 3).

x squared + x) squared - 14 (x square + x) + 24 (to process) The square of x + x is regarded as a term t, and the original formula = (t-2) (t-12) == (x 2 + x-2) (x 2 + x-12).

ab+ac—b squared—2bc—c squared (process) (ab+ac)-(b2+bc)-(bc+c2)a(b+c)-b(b+c)-c(b+c).

a-b-c)(b+c)

x—y) squared—4(x-y+15) process.

Think of x-y as t, and the original formula = t 2-4t-60

Use cross multiplication = (t+6)(t-10).

x-y+6)(x-y-10)

ax+ay+bx+by—a—b process.

a(x+y)+b(x+y)-(a+b)

a+b)(x+y)-(a+b)

a+b)(x+y-1)

Ten x square (x square + y squared) — 20y four times.

Cross multiplication, -20y quadratic to -4y 2*5y 2 original = (x 2-4y 2) (x 2 + 5y 2) (x-2y) (x + 2y) (x 2 + 5y 2).

1 -y(x-3)(x-5)

2 (x to the nth power - 11) (x to the nth power +6) 3 (3 to the m power + 9/2).

4 (a to the m power + 3x) (a -2y).

5 (x+3-y)(x+3+y)

6 (x²+x)²-14(x²+x)+24((x²+x)-2)((x²+x)-12)(x+2)(x-1)(x+4)(x-3)

7 ab+ac-b²-2bc-c²

a(b+c)-(b+c)²

a-b-c)(b+c)

8 (x-y)²-4(x-y+15）

x-y)²-4(x-y)-4*3*5

x-y)-2*5)((x-y)+2*3)(x-y-10)(x-y+6)

9 ax+ay+bx+by-a-b

a(x+y)+b(x+y)-(a+b)

a+b)(x+y)-(a+b)

a+b)(x+y-1)

10 x²(x²+y²)-20(y²)²

x²) x²y²-20(y²)²

x²+5y²)(x²-4y²)

x²+5y²)(x-2y)(x+2y)

y(x-3)(-x+5)

x^n-11)(x^n+6)

3^m + 9/2)^2

I didn't get it. x^n+3-y)(x^n+3+y)

x^2+x)^2-14(x^2+x)+24 = (x^2+x)^2-(2+12)(x^2+x)+(2*12)=(x^2+x-2)(x^2+x-12)=(x+2)(x-1)(x+4)(x-3)

ab+ac-b^2-2bc-c^2 = a(b+c) -b+c)^2 = (a-b-c)(b+c)

x-y)^2 - 4(x-y+15) = (x-y)^2-4(x-y)-60 = (x-y)^2-(10-6)(x-y)-6*10 = (x-y-10)(x-y+6)

ax+ay+bx+by-a-b = a(x+y-1) +b(x+y-1) = (a+b)(x+y-1)

What's the last line?