Mathematics.. It is known that x 1, y 10 and x 2, y 1 and x 2, y 7 are all solutions of y ax 2 bx c,

Let's substitute it. 1. x=1，y=10

So 10=a+b+c

2. x=2，y=-1

So -1=4a+2b+c

3. x=-2，y=7

So 7=4a-2b+c

6=8a+2c

Simplification 3 = 4a + c

So c = 3-4a

2 20=2a+2b+2c ④

21=2a-c will bring in.

So -21=2a+4a-3 a=-3 is known by c=15 to bring a=-3 c=15 into the equation b=-2

So. a=-3 b=-2 c=15

Bring x=1, y=10, x=2, y=-1, x=-2, y=7 into y=ax 2+bx+c respectively to form a system of equations, and then use the knowledge you have learned to solve the system of equations! For example: (bring the upper value in).

Curly braces 10 = a + b + c

Curly braces - 1 = 4a + 2b + c

Curly brace 7 = 4a-2b+c

With - remove a and c

Solve b=-2

Jian substitutes b with and , with -

Solve a=-3

Substituting a and b into , we can solve c=15

The final analytic formula is y=-3x 2-2x+15

Substituting x=1 and y=10 into y=ax 2+bx+c, we get 10=a+b+c——Eq. 1

Substituting x=2,y=1 into y=ax 2+bx+c, we get -1=4a+2b+c—Eq. 2

Substitute x=-2,y=7 into y=ax 2+bx+c, and get 7=4a-2b+c—Eq. 3

Subtract Eq. 3 from Eq. 2 to get -8=4b, i.e. b=-2

Subtract Eq. 2 from Eq. 1 to get 11=-3a-b, substitute b=-2 into A=-2, and substitute a=-3 and b=-2 into Eq. 1 to get c=15

Divide it into three substitutions to obtain a system of ternary equations, and you can solve it.

Analysis: The deformation x2+y2+xy-x+y+1=0 gets the wild plum to x2+(y-1)x+y2+y+1=0, and regards it as the absolute grasp of x's unary quadratic equation, because x has a value, according to the meaning of 0, that is, (y-1) Song Hongchi 2-4(y2+y+1) 0, the deformation has (y+1)2 0, using the property of non-negative numbers to obtain y+1=0, and the solution obtains y=-

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

x(x+y)〔(x-y)-(x+y)〕

x(x+y)(-2y)

2xy(x+y)

Substitute x+y=1, xy=-1 2 into -2xy(x+y) to get a dry cavity cavity oak shed:

2xy(x+y)

x|=1,|y|=2, we get x= 1, y= 2, and |x+y|=x+y, the reputation of the blind god defeated x+y>0, then x=1, y=2, or qingjiao x=-1, y=2, so x-y=-1 or -3

Solution: Put when x=-1, y=10; When x=1, y=4;When x=2, y=7

Respectively, let the sales of friends annihilate into the good slippery rush:

10=a-b+c

4=a+b+c

7=4a+2b+c

Solve the system of equations and get:

a=2，b=-3，c=5

So: y=2x 2-3x+5

When x=-1, y=10=a*(-1) 2+b*(-1)+cx=1, sell and undo y=4=a1 2+b*1+cx=2, y=7=a2 2+2b+c

Solving the system of equations yields a=2, b=-3, c=5

The shed guessed to the sedan y=2x 2-3x+5

The common factor of x(x+y)(x-y)-x(x+y) 2 is x(x+y), and the common factor is obtained

Original formula =x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=-2xy(x+y), substituting the known condition x+y=1, xy=-1 2.

Original = (-2) * (1 2) = 1

The second question is 1 2a3b+a2b or 1 (2a3b+a2b).

x(x+y)(x-y)-x(x+y)²

x(x²-y²)-x(x²+y²+2xy)=x³-xy²-x³-xy²-2x²y

2xy²-2x²y

2xy(x+y)

1. Question 2: Are you sure the question is complete?

x+y=2

square x +2xy+y =4

then x -2xy + y = 4-4xy

xy=1 is known

So (x-y) = 4-4*1=0

Therefore ix-yi=0

Knowing x+y=3, xy=-10, find (1)x-xy+y; 2）｜x-y ｜

Solution 1: x+y=3, xy=-10, x,y is the root of the equation z-3z-10=(z-5)(z+2)=0: z =x=-2, z =y=5

or z = y = -2; z₂=x=5. ∴x-y│=│-2-5│=│5+2│=7

1) x -xy=y may be wrong, it should be x -xy+y or x -xy-y

x²-xy+y²=4+10+25=39;x -xy-y = 4 + 10 - 25 = 11 or 25 + 10 - 4 = 31

Solution 2: (x+y) =x +2xy+y =x -20+y =9

Therefore x +y = 29 , x -xy + y = 29 - (-10) = 39

x-y)²=x²-2xy+y²=29+20=49

Hence x-y =7