The math problems of high 1 do not seem to be difficult, and they are willing to solve them

1.(1) f(x)=x analytical, my favorite function method. It profoundly encompasses the various forms of functions.

2) Mapping a{x, x belongs to r}b{y, y belongs to r}a->b f:y=x Basic function definition, but not intuitive.

3) Then there are the images. It is the most intuitive, but it is not as convenient as the analytical calculation in terms of quantitative calculations.

x>=2 y=2-x x<2

The image, symmetrical with x = 2, draw two symmetrical straight lines on each side.

f(3)=3-2=1

f[f(3)]=f(1)=2-1=1

3.(1) No, the definition domain is not the same, f(x) is r, g(x) is x≠0

2) Yes, it doesn't matter what letter the function uses as a variable. h(a)=3a+2 m(d)=3d+2, both are a function.

3) No, the previous definition domain is r because x 2 is not negative, and the later definition domain is x 0

4.(1)x 2-5x-6=(x-6)(x+1) 0 then the domains are x 6 and x -1

2) x+1 0 x -1 x-6 0 x 6 takes the intersection then the domain is defined as x 6

3)2-x^2>0 -√2< x< 2 and x≠0< p>

It's too easy, it's just that I'm too lazy to type!

aub=

Does a reversal indicate the complement of A, yes, just go down, if not, you don't need to look down.

a inverts ub=, cr(a inverts ub)= , which indicates an empty set.

cra) inverted u

b=aub=

au（crb）=φ，

Let a and b be two unequal positive numbers, and a 2-b 2=a 3-b 3, and verify 14ab

So ab<(a+b) 2 4

So ab<(a+b) 2 4

So (a+b) 2-ab>(a+b) 2-(a+b) 2 4=3(a+b) 2 4

Thus a+b>3(a+b) 2 4

Solution: 0a+b

Solve a+b>1 or a+b<0 (rounded) (3) from (2), (3) get 1

Solution: Let x=x 2, then:

......... by f(x)-(1 2)f(x2)=x2Hungry .........Gotta :

f(x/2)-(1/2)f(x)=(x/2)^2………So by rotten quietly stove x(1 2)+ get:

f(x)=(3 Yunpai2) x 2

The range is r, that is, the part of the function value greater than or equal to zero for ax 2+ax+1>0 should be taken.

So ax 2+ax+1 is a>0 and the discriminant is greater than or equal to zero.

Only in this way can you get all the parts of ax 2+ax+1>0.

Solution: Let the projection point be n(x,y,z), then.

nm0=(x-1,y-1,z), and the normal vector of this plane is (1,1,-1), so x-1=y-1=-z and x+y-z=3

The solution is: x=y=4 3, z=-1 3

The projection points are (4 3, 4 3, -1 3).

Set to b(a,b,a+b).

It turned out that the point was a

ab=(a-1,b+1,a+b-0)

The plane normal vector m=(1,1,-1).

So a-1=b+1, -1-b=a+b-0

a=1,b=-1

Bring it back and have b coordinates.

by f(xy) = f(x) + f(y).

If f(x)=f(x y*y)=f(x y)+f(y), then f(x y)=f(x)-f(y) is proven.

by f(xy) = f(x) + f(y).

f(9)=f(3*3)=2f(3)=2*1=2, so f(x)>f(x-1)+2 =f(x-1)+f(9)=f[9(x-1)].

From f(x) is a monotonic addition function defined on a positive integer, we get x>9(x-1) and x<9 8