Asking for a math problem in the first year of junior high school is urgent, urgent

First, 2-ab +(b+1)2=0 can push out b = -1 a = -2

As prompted, consider b = a + 1, then 1 (a-x)(b-x) = 1 (a-x)(a-x+1) = 1 (a-x) -1 (a-x+1).

Here x refers to the number in the range 0-2010.

So the original formula is decomposed as -> 1 a - 1 (a+1) +1 (a-1)-1 a + 1 (a-2) -1 (a-1).

1/(a-3) -1/(a-2) +1/(a-2010) -1/(a-2009)

The positive terms are 1 a + 1 (a-1) +1 (a-2) +1 (a-2010).

The negative term is - 1 (a+1)-1 a - 1 (a-1).

After the intermediate offset, it is 1 (a-2010) -1 (a+1) = - 1 2012 + 1 = 2011 2012

The idea is this, the landlord can check for himself if there is a problem.

Because |2-2b|Greater than or equal to 0, so b=1(b+1) squared is greater than or equal to 0, so b=-1

So this question was wrong from the beginning.

2-2b +(b+1)2=0 Error.

Find 1 ab+1 (a-1)(b-1)+1 (a-2)(b-2)+....The value of 1 (a-2010) (b-2010) (hint 1 n(n+1) = 1 n-1 n+1 less condition.

5. When A moves to B, the length of one AB is moved, and when B moves to A, the length of another AB is moved, so the length of 3 wooden strips = (20-5) 3=5

[5 k-n]= [5 k] n (n is a positive integer) when k 2, xk=xk-1+1-5= xk-1+1-5xk-1+1-5= xk-1+1-5(-1+2)= xk-1-4 In the same way, yk=yk-1+[5 k-1]-[5 k-2]= yk-1+1 When x=1, x1=1, y1=1;

When x=2, x2=1-4=-3 , y2=1+1=2;

When x=3, x3=1-4 2= -7 , y3=1+1 2=3;

When x=4, x4=1-4 3= -11 , y4=1+1 3=4;

When x= n, xn=1-4 (n-1)=5-4n , yn=1+1 (n-1)=n;

When x= 10, x10=5-4 10= - 45, y10=10;

When x= 2009, x2009=5-4 2009=-8031, y2009=2009

x(k)=x(k-1)+1-5[t[（k-1）/5]-t[(k-2)/5]]，y(k)=y(k-1)+t[(k-1)/5]-t[(k-2)/5].

I guess so

t[(k-1) 5] is equation (1).

t[(k-1) 5] is equation (2).

Let n be a positive integer.

2t[(k-1)/5]-t[(k-2)/5]=1x=6-5=1

y=1+1=2

When n5+1t[(k-1) 5]-t[(k-2) 5]=0x=k-5n

y=1+n