A math fill in the blank question. There is an answer, but what you want is analysis.

Odd function: f(-x) = -f(x).

Bring in the expression satisfied: -f(x)+f(x+2)=0

Yes: f(x)=f(x+2).

Defined by the period function, f(x) minimum positive period is 2, so 4 is its period.

Take (a,b) on y=f(x). If y=f(x) is symmetrical with respect to x=0, then when there is (-a,b) is also on y=f(x).

i.e. f(a)=b=f(-a).

Let a=x-2 (Note: X here is also any drop...) i.e. f(x-2) = f(2-x), this condition is known.

If y=f(x) is symmetrical with respect to x=2, then the distance to 2 on the x-axis (i.e., |x-2|Equal points have equal function values. i.e. f(2-x) = f(2+x), which is equivalent to y=f(x+2) as an even function.

If you take a, and take the point (2+a,f[(2+a)-2]) on y=f(x-2), then its symmetry point with respect to x=2 is (2-a,f[(2+a)-2]), i.e., (2-a,f(a)); Take x=2-a, and bring in the function y=f(2-x), y=f[2-(2-a)]=f(a), which shows that the point (2-a, f(a)) is on y=f(2-x). Due to the arbitrariness of place a, all the symmetrical points on y=f(x-2) with respect to x=2 are on y=f(2-x).

In the same way, it can be proved that all the points on y=f(2-x) with respect to x=2 symmetry are on y=f(x-2).

That is, y=f(x-2) is symmetrical with y=f(2-x) with respect to x=2.

This problem is very simple, because the problem is set simply, so the trick to doing this kind of problem is to make a function that meets the conditions and then try it on your behalf.

Analysis: 1. Odd function, then f(-x)=-f(x), that is, -x=x+2, x=-1, that is, when x=-1, 2f(1)=0, f(1)=0;According to the property of the odd function about the symmetry of the origin, f(-1)=0, bring in x=0,1 in turn, f(0)=-f(2), f(-1)=f(3)=0, draw the image, the specific value of f(0) is arbitrarily determined, it can be seen that it is a periodic function with a period of 4;

2. Let x-2=t, then f(t)=f(-t) satisfies the condition of the even function, and the property of the even function is symmetry with respect to the y-axis (x=0);

3. Let f(x)=x 2, then f(x+2)=(x+2) 2, substituting x=-1,0,1,2,3 can get symmetry about x=2;

4. In the same way, let f(x)=x, f(x-2)=x-2, f(2-x)=2-x, and substitute 0,1,2,3,4 to obtain symmetry about x=2.

A = two-thirds times b, so b = a divided by two-thirds = three-thirds a, i.e. b is three-thirds of a.

Make a circle with a as the center of the circle, with a radius of 2 3 3

The arcs that intersect the square are what is sought.

When finding the length, start with the square where the green sector is located.

Finding tan shows that the two angles next to the central corner of the green fan are both 6, so the central angle of the green fan is 6

In the same way, the blue central angle is 3, and the red is 4

Because arc length = radius of the central angle.

So the length is: 2 3 3 ( 6 + 3 + 4) = 3 2

Putting the square, from the length of 2 root number 3 3 can be calculated that the set with the distance from each face is actually an arc, and the radian is 30 degrees, so the length of the curve is (6 * 30 360) *2 * (2 * root number 3 3) * = (2 * root number 3 3) *

Several 15 cm by 20 cm floor tiles can be completely put together to create a square with an area of at least (3600) square centimeters.

From the question, f1ab is an isosceles right triangle and a is a right-angled vertice.

then, let af1 ab m

bf1＝√2m

It can be obtained from the nature of hyperbola.

af2＝af1－2a＝m－2a

bf2＝bf1－2a＝√2m－2a

then, ab af2 bf2 (1 2)m 4a again, ab m

So, (1 2) m 4a m

i.e., 4a 2m

So, 16a 2m

F1af2 is RT

f1f2＝2c

f1a mf2a m 2a (1 2 2) m by the Pythagorean theorem, f1f2 f1a f2a then, (2c) m (3 2 2) m i.e., 4c (5 2 2) m

So, 16c (10 4 2)m

e²＝c²/a²

16c²)/16a²)

5 Socks imitation 2 2

3 solution: x+y+z=5 because.

x+2y+3z)+(4x+3y+2z)=10+155(x+y+z)=25

x+y+z=5

4 Solution: Because half of the remaining half plus 1 for.

1/2*(1/2ab-1)+1

ab/4+1/2

The remaining books are:

ab-( 1/2 ab+1)-(ab/4+1/2)=ab/4-3/2

3.(x+2y+3z) +4x+3y+2z) = 10 + 15 = 25

5(x+y+z) = 25

x+y+z = 5

4.After donating to the welfare home, the remainder is: ab 2

Donate the remaining half to the school for the deaf and dumb, and there is left: ab 4

Then give a copy to the school for the deaf, and eventually left: ab 4 - 1

Question 1: 5

Add the two formulas to get 5x+5y+5z=25, and then divide both sides by 5 at the same time Question 2: quarter ab + half.

The remaining half of the ab-1 of the donation home, and then the second half (half of the ab-1) + 1 is simplified.

x+2y+3z=10

4x+3y+2z=15

The sum of the two formulas.

5x+5y+5z=25, then.

x+y+z=5

There is a total of ab books, donated to the welfare home ab 2 + 1, the rest ab 2-1, donated to the deaf school ab 4-1 2 + 1, that is, ab 4 + 1 2, so the remaining ab 2-1-ab 4-1 2, that is, (ab-6) 4