A math problem If 1 A 2B 3 2 2A B 6 find A ? B How to solve a problem like this?

From 2 2a-b 6 we know -6 b-2a -2 (3), i.e., -12<2b-4a<-4

Add it to Eq. (1) to get -13<-3a<-1 to get a (1 3 , 13 3).

Bringing 2a into equation (3) gives b ( 16 3 , 20 3 ) Note that inequalities can only be added, not subtracted, multiplied by positive invariant signs, multiplied by negative variable signs.

Method 2: Linear programming (draw your own plot).

Columnable inequality groups: a-2b -1 solution a 2b+1 b (a+1) 2a-2b 3 a 3+2b b (a-3) 22a-b 2 solution a (2+b) 2 b 2a-22a-b 6 a (6+b) 2 b 2a-6(a+1) 2=2a-2, a=5 3

2b+1=（2+b）/2，b=-2/3

a∈5/3，b∈-2/3

Upstairs is very correct, how do you know that 4038090 is 2009*2010... Strong!

1－2＋3－4＋5－6＋.199 200 is -1 for each group of two

So the original formula = (-1) (200 2) = -100 x 1 y

x-1=0，x=1

y, y=so x+y-1=

Have fun.

Question 1: (1-2)+(3-4)+. 199-200)=-1-1-1-..1=-100

Question 2: Because |x-1|Greater than or equal to 0, |y+ is greater than or equal to 0 and because x 1 y

So x=1. ，y=

So x y 1=

Question 1: —100, Question 2: The equation is only true if the absolute value is zero, so x is 1, y is —so the answer is —

From the function y=1 2x+2, we can get the point a(-4,0) and the point b(0,2), and we can see from the graph that the height of pao is y, so s=(1 2)|ao|y=2y=2（1/2x+2）=x+4 （-4

s=1 2*4*y, y=1 2x+2, just bring the second one in, this is a primary school topic or a junior high school topic.

By 1 (n*(n+1))=1 n-1 (n+1), you can prove it yourself.

Original formula = 1 1-1 2+1 2-1 3....1/2011-1/2012

You can split 1 (1 2) into 1 1 -1 2 and so on.

According to 1 (n*(n+1))) = (1 n)-(1 (n+1)).

In the end, 1 - (1 2012) = 2011 2012

The answer is in the picture of the lack of Senxun, I don't know if you can understand this spring fierce, click** to view the big picture.

Because the difference between each denominator of the original formula is 2, such as 19 and 21, 21 and 23, etc., 1 19-1 21 is exactly equal to lead wax (2 19 21), so after simplification, it should be multiplied by the open crack 1 2, and the rest of the equations are the same principle.

solution, divide both sides of 9b +2008b + 5=0 by b.

Then 5*(1 b) +2008*(1 b)+9=0

Think of a,1 b as the root of the equation 5x 2008x+9=0.

Then: a (1 b) = 9 5

a/b=9/5.Pick B.

9b²+2008b+5=0

Divide by b on both sides

5(1/b)²+2008(1/b)+9=05a²+2008a+9=0

So a and 1 b are the root of the equation 5x +2008x +9 = 0 by Vedic theorem.

a*1/b=9/5

So a b = 9 5

Obtained from the question: a=(-9-5a) 2008, b=(-5-9b) 2008

Divide the two formulas to obtain: a b = (9+5a ) (5+9b ) = >5a + 9ab = 9b + 5a b

5a(ab-1)-9b(ab-1)=0=>(5a-9b)(ab-1)=0

5a-9b=0, ab-1=0 (rounded).

So a b = 9 5

The condition should be ab≠1, right?

Otherwise, you can't ask for it.