One math problem is compulsory in high school, and one math problem is compulsory in senior high sch

Solution: a= x|x -5x+q=0 q 25 4x -5x+q=0 can be reduced to (x-5 2) -25 4+q=0x-5 2) =25 4-q

q≤25/4 25/4-q≥0

So the guarantee (x-5 2) = 25 4-q 0 is constant.

x can get 1, 2, 3, 4, 5

So the complement UA is an empty set.

There is only 1 in the set

Hopefully, thank you.

Solution: f(2)=1,f(x)=x (ax+b) 2=2a+b

f(x)=x

f(x)=x/(ax+2-2a)=x

ax^2+(1-2a)x=0

There is a unique solution. △=1-2a)^2=0

1-2a=0,a=1/2

b=1∴f(x)=2x/(x+2)

Hope it helps

1 All 1Let x=y=1

Substituting f(x y)=f(x)-f(y).

f(1)=f(1)-f(1)=0

The inequality f(x+3)-f(1 3) 1

f（x+3)-f(1/3)＜f（6）

f(3x+9)＜f（6）

Because the function f(x) is defined as an increasing function over (0,+infinity), 3x+9<6

and then according to the definition domain.

x+3>0

So the answer is -3

Let x=y=1, we get f(1)=0

The original inequality is f(3x+9)0, so x>-3, so -3

(1) Let y=

2) Let x=1, f(1 y)=f(1)-f(y)=-f(y), then the original inequality becomes: f(x+3)+f(3)0, so x>-3, so -3

by a=b

m+d=mq and m+2d=mq (q squared).

The common factors m(1-q)=d and m(1-q2)= - 2d, and because m(1-q2) = -2d, it can be reduced to m(1-q)(1+q)= -2d

Simultaneous m(1-q) = d m(1-q)(1+q) = -2d to obtain 1+q= -2

So q= -3

Since a=b, then m+d=mq, m+2d=mq 2, solving these two equations gives q value of 1 and d value of 0

(q(q squared)-1)m=2d

q-1)m=d

(q(q-squared)-1)m=2(q-1)m

Approximately go m, and q(q squared)-2q+1=0

The solution is q=1

The individual elements are equal.

Make a list of the two equations.

Eliminate m and dq = 1

1 Piecewise function, substituted by the corresponding analytic formula, f(-1)=f(-1+3)=f(2)=f(2+3)=f(5).

5-4=1,2,∵f(x)=cx/(2x+3),∴f[f(x)]=c^2x/(2x+3)]/2cx/(2x+3)+3]

c^2x/(2cx+6x+9)

x, i.e., c 2 = 9, and 2c + 6 = 0, c = -3

1. f(-1)=f(3-1)=f(2)=f(3+2)=f(5)=5-4=1.(mainly based on the domain defined).

2.[c cx (2x+3)] 2 cx (2x+3)+3]=x, that is, the whole formula of x in f(x) is replaced by f(x), and both sides are divided by x, (2c+6)x+9-c 2=0, (the square of c), since it has no relationship with x, 2c+6=0, 9-c 2=0, thus c=-3

Solution: Except for the absolute value -6-8(1)a is positive due to -1 so a=-4 (the assumption is not true).

2) When a is negative, the same solution as above a=-4

Let f(x)=x -1 3 show monotonicity: when x 0, f(x) decreases with the increase of x;

When x<0, f(x) increases with the increase of x.

A+1>0,3-2A>0,A+1>3-2ATo sum up, the value range of A is (negative infinity,-1) (2 3,3 2) What, negative infinity is too lazy to play, you should know.

Also, I'm also a freshman in high school. The idea must be correct, and I hope there are no mistakes in the answer.

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