Friends, please help!! A math problem

x2-x=13,y2-y=13

x2-x=y2-y

x2-y2=x-y

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

It is known that x is not equal to y, then x-y is not equal to 0

So about (x-y): x+y=1

From the meaning of the problem, we can know that x and y are the two solutions of the equation x -x=13.

This should be easy to understand, a quadratic equation has this property.

Then you don't need to rush to solve the equation, because the problem requires you to calculate the value of x+y, so you can use Veda's theorem:

Suppose the two solutions in the quadratic equation ax +bx+c=0 (a is not equal to 0) are x1 and x2

Then x1+x2= -b a x1*x2=c a, so x+y=-(-1) 1=1

x+y=1x2-x=13 ……1）

y2-y=13 （2）

Eq. (1)-Eq. (2), x2-x-(y2-y)=0 (x2-y2)-(x-y)=0

x+y)(x-y)-(x-y)=0 (x-y)[(x+y)-1]=0 Because x is not equal to y, x+y-1=0 i.e. x+y=1

Subtract the formula to get (x+y)*(x-y)=x-yx is not equal to y, so x-y is not equal to 0

Divide both sides by (x-y) at the same time.

The idea of x+y=1 is the same.

Add and subtract the two formulas given and simplify them.

Use the squared difference or sum of squares formula.

Evaluation, of course, can also be made easier by Vedic theorem.

Solution: x+y=1

Rationale: From the known, x 2-x=13, y 2-y=13 - obtained. x+y）(x-y)-(x-y)=0x-y)(x+y-1)=0

Because x is not equal to y, x+y=1

x,y is the square of the equation x minus x is equal to the two roots of 13, so x+y=1

According to the formula of the sum of the two roots of the quadratic equation, we know that x+y=13

Exactly as .

Approximate solution is sum.

The image of the function f(x) defined on r is symmetrical with respect to the point (-3 4,0), then there is.

f(x)=-f(-x-3 2), and for any real number x there is f(x)=-f(x+3 2).

f(x)=f(-x)，f(x)=f(x+3)。So f(1)=f(2)=f(—1)=1, f(3)=f(0)=-2, f(1)+f(2)+f(3)=,f(6)=-2,f(4)+f(5)+f(6)=0.

f（1）+f（2）+f（3）+…f（2010）+f（2011）=f（2011）=f（1）=1