Can you help me with this high school math problem?

0=f(x+2)=f(x) means that the image is a function of 2, so f(2010)=f(0+2*1005)=f(0)=0

Because the function is an even function, f(-2011)=f(2011)=f(1+2*1005)=f(1)=2 1-1=1

So: f(2010)+f(-2011)=0+1=1If you don't understand it, you don't understand it!

f(x) is an even function, f(x) = f(-x).

f(-2011)=f(2011)

f(2010)=f(2+2008)

When x>=0, there is f(2+x)=f(x): f(2+2008)=f(2008).

In the same way, the cycle follows: f(2010)=f(0) f(2011)=f(1), then f(2010)+f(-2011)=f(0)+f(1) and x is greater than or equal to 0, and when it is less than 2, f(x)=2 (x-1)f(0)=1 2 f(1)=1

obtained: f(2010)+f(-2011)=f(0)+f(1)=1 2+1 =3 2

f(2010)+f(-2011)=f(2010)+f(2011)=f(2008)+f(2009)=……

f(0) + f(1) = 2 to the power of 0 + 2 to the first power of 2 = 3

The power of 2 x can be represented as 2 x in this way

To do this kind of problem, you must draw a picture, and it is clear at a glance.

It is not difficult to investigate the properties of the sine (co)sine function - periodicity and parity.

The period of the sine and cosine functions is 2, and a semicircle is one.

Solution: f(log (x-3)) + f(log x) 0, to f(log (x-3)).

f(log x) and f(x+1) are odd functions f(x+1)=-f(-x+1).

f(log₂x)=f((log₂x-1)+1)=-f(1-log₂x+1)=-f(2-log₂x)

f(log₂(x-3))＞f(2-log₂x)

f(x) is a monotonically decreasing function on [1,+ f(x+1) in.

0,+ is a monotonically decreasing function and f(x+1) is an odd function

f(x+1) is also a monotonically decreasing function on [0,+ f(x) is a monotonically decreasing function on (- 1] has log (x-3)) 2-log x,x-3)x 2 ,x -3x-4 0,(x-4)(x+1) 0,get:

1 x 4 x-3 0, x 0 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 0 x 0 x 0 x 0 x 0 x 3 x 3 x 3 x 3 x 3 x 0 x 0

Please refer to it. Equations containing unknown quantities are equations, mathematics first developed in counting, and about numbers and unknowns are combined through addition, subtraction, multiplication, division, and idempotency to form algebraic equations: unary equations, unary quadratic equations, binary linear equations, and so on.

However, with the emergence of the concept of functions and the introduction of function-based differentiation and integration operations, the scope of equations has become more extensive, and unknown quantities can be mathematical objects such as functions and vectors, and operations are no longer limited to addition, subtraction, multiplication and division.

Equations occupy an important place in mathematics and seem to be an eternal topic in mathematics. The emergence of equations not only greatly expanded the scope of mathematical applications, enabling many problems that could not be solved by arithmetic problem solving, but also had a great impact on the progress of mathematics in the future. In particular, many major discoveries in mathematics are closely related to it.

The unknowns in equations can appear in the equations, integers, radicals, trigonometric functions, exponential functions and other elementary functions in the equations.

In secondary school, when you encounter problems with solving equations, in general, you can convert the equation into an integral equation; Generally, it is a problem of converting imaginary models into one-dimensional quadratic equations, or solving multivariate systems of one-dimensional equations.

The formula for finding the root of a quadratic equation.

Since mathematics has changed from constant numerical delay to variable mathematics, the content of equations has also been enriched, and more concepts and operations have been introduced into mathematics due to differences, thus forming more equations. The development of other natural sciences, especially physics, has also directly put forward the need for equation solving, providing a large number of research topics.

First of all, p(b a) means the possibility of b in the case of satisfying a.

Then, a is the positive imitation of the description The first time is an odd number, there are a total of 13579 odd numbers, so C51, and the second time there are eight left, just take one of them to C81, so the possible number of events of A is C51*C81=40

If you look at event b, you need to satisfy event a first, so you must get an odd 123579 first time. Now there are two situations that happen: the first is that the number you get for the first time is not only an odd number, but also a multiple of 3, there are only 39 two, so c21, and then the second time you get one of the remaining two multiples of 3 is also c21, so c21*c21; The second is that the first time you get an odd number but not a multiple of 39, so the first time you can only take it from 157, it is c31, and the second time you get it from the multiple of the remaining 3, which is 369, which is c31, so the second case is c31*c31.

The two kinds of year-old fiber cases are added together with C31*C31+C21*C21

Actually, I don't think the denominator is 3 because m and the two points don't coincide.

That is, the value of this formula should be between three-fifths and one-fifths, three-fifths can be taken, and 1 cannot.

The tangent of this question is to seek guidance.,Ask for guidance in high school.,If you say you can't ask for guidance.,Then you go back and review it yourself.。

y=x+2/x

y'=1-2/x^2

Straight line l: y-3=-(x-1), that is, x+y-4=0 origin o to straight line l distance = 4 2 = 2 2