It is known that the function f x defined on 0, positive infinity is an increasing function, and i

Let x1=x2=4 obtain: f(16)=2f(4)=2 first meet the requirements of the defined domain: x+6>0, x>0 obtains: x>0;

f(x+6)+f(x)=f(x 2+6x), so the original inequality is f(x 2+6x)>f(16).

Because f(x) is an increasing function.

So, x 2+6x>16

Solution: x -8 or x 2

And because the definition domain requires x 0

So: x 2

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f(x1x2)=f(x1)+f(x2), and f(4)=1f(4*4)=f(4)+f(4)=2

f(x+6)+f(x)＞2

The function f(x) defined on (0, positive infinity) is an increasing function.

x+6>0,x>-6,x>0

f(x+6)+f(x)>2

f(x*(x+6))>f(16)

So, x 2+6x>16

x<-8 or x>2

So the synthesis gives x>2

From the problem, f(16)=f(4)+f(4)=2 and f(x+6)+f(x)=f(x2+6x), i.e. f(x 2+6x) f(16).

f(x) at (0,+ is an increasing function.

x^2+6x-16＞0

Solution x (-8)u(2,+

x 0 gets x (2,+

Let x=y.

f(x x) = f(1) = f(x) - f(x) = 0f(x+3) - f(1 3) = f(3x+9)<2 by f(x y) = f(x)-f(y).

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

f(6)+f(6)=2

i.e. f(36 6) + f(6) = 2

f(36)-f(6)+f(6)=2

f(36)=2

The original inequality can be reduced to.

f(3x+9)0

The set of solutions to this inequality is .

f(x) is an increasing function defined on 0 to positive infinity, and for all x,y>0, satisfying f(x y) = f(x)-f(y).

Let x=y=1

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

f(1)=0

f(6)=1

Let x=36, y=6

f(36/6)=f(36)-f(6)

f(6)=f(36)-f(6)

2f(6)=f(36)

f(36)=2

f(x+3)-f(1/3)<2

For all x,y>0, f(x y)=f(x)-f(y)f[(x+3) (1 3)]<2=f(36)f(3x+9)0

So the solution of the inequality is.

0

f(x y) = f(x) -f(y) let x=1 and y=1 then f(1) = f(1) -f(1) so f(1) = 0

f(x+3)-f(1 3)<2=2f(6)f(x+3)-f(1 3)-f(6)f(x 2+3 2) Because this function is an increasing function, x 2+3 2<6x<9

Since x is defined on (0, positive infinity), the value of x can range from (0,9).

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

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

f(6)+f(6)=2

i.e. f(36 6) + f(6) = 2

So f(36)=2

The original inequality is reduced to.

f((x+3)/(x-3))0*****==>x>-3(x-3)>0*****==>x>3

x+3) (x-3)<36==>x>111 35, so x>111 35

I hope it can help you, and I wish you progress in learning, and don't forget to adopt!

f(x,y)=f(x)+f(y)

f(x)+f(x-3)=f(x^2-3x)f(2)=1

2=2f(2)=f(2)+f(2)=f(2*2)=f(4)y=f(x) is in (0.).+.

f(x)+f(x-3)>2 is equivalent to.

f(x2-3x)>f(4) are equivalent.

x^2-3x>4

x^2-3x-4>0

x-4)(x+1)>0

x>4 or x<-1

x>0x>4

See the picture Please don't forget to adopt Happy Learning.

1.∵f（x/y）=f（x）-f（y）

Let x=y=1

f(1)=f（1/1）=f（1）-f（1）=0∴f（x-1）＜0

i.e. f(x-1) f(1).

and f(x) is an additive function defined on (0, positive infinity) solution f(x-1) f(1).

i.e. solution x-1 1

Solution x 22∵f（2）=1

Solution f(x+3)-f(1 x) 2

i.e. f(x+3)-f(1 x) 2f(2) f(x+3)-(f(1)-f(x)) f(2)+f(2) f(x+3)-(0-f(x)) f(2)+f(2) f(x+3)+f(x) f(2)+f(2) f(x+3)-f(2) f(2)-f(x)-f(2)-f(x)(x+3) 2) f(2 x).

i.e. solution (x+3) 2 2 x

It can be solved with x -4 or 0 x 1

f(x) defines the domain as (0, positive infinity).

In summary, the solution is 0 x 1

(1) Let x=ythen there is f(1)=0

And because f(x) is an increasing function on (0, positive infinity), when 0 solves the inequality f(x-1)< 0, i.e. 0(2), we can know that f(1)=0, and in f(x y)=f(x)-f(y), we get f(1 y)=f(1)-f(y), and f(x)=-f(1)@

f(2)=1,f(1)=0 (let x=1,y=2 be substituted into f(x y)=f(x)-f(y)) to get f(1 2)=-1,, and then x=2,y=1 2 to f(x y)=f(x)-f(y) to get f(4)=2

The inequality can be changed to f(x+3)+f(x)<0 by the @ formula, since f(x) is an increasing function, so f(x+3)+f(x) is also an increasing function (the addition of two increasing functions) so that f(x)=f(x+3)+f(x), then f(1)=f(4)+f(1)=2, so the solution of the inequality f(x)<0 is 0

(1) x-1>0, let x=y=1, f(1)=0, f(x-1)<0=f(1), because the multiplication function obtains, x-1<1, so 10,1 x>0, (x+3)x<4,0

(1) m=n=1 f(1)=f(1)+f(1) f(1)=0 1 is the zero point.

2) There is a problem with the narrative, as mentioned earlier when x 1, f(x) 0, but f(2) = 1 2>0

Should it be x1 when f(x) > 0

f(2)=1/2

f(4)=f(2)+f(2)=1

And for any x>4 there is f(x)=f(4)+f(x 4) and because f(x 4)>0 there is f(x)>f(4)=1

Any 00 so the solution with f(x)1 is ax+4>4 ax>0

If a>0 is solved as x 0,

If a=0 or a<0 there is no solution.

(1) f(1*1)=f(1)+f(1)=2f(1)=f(1) can be obtained f(1)=0

2) For any m 0, f(m*(1 m))=f(1)=f(m)+f(1 m)=0 f(m)=-f(1 m) can be known

For any x1, x2 0, let x11, f(x) < 0, so f(x2)-f(x1)=f(x2)+f(1 x1)=f(x2 x1)<0, i.e., f(x) is monotonically decreasing.

f(2)=1 2 f(4)=f(2)+f(2)=1 inequality is converted to ax+4<4 ax<0 a<0 because x>0 is a0

It should be understandable. If you have any questions, get in touch again.

Let f(x) be an increasing function defined on (0, positive infinity), and for any x, y belongs to (0, positive infinity), there is f(xy) = f(x) + f(y).

1) Verification: f(x y) = f(x)-f(y)f(y y)=f(y)+f(1 y).

f(1)=f(y)+f(1/y)

f(x)+f(1)=f(y)+f(x)+f(1/y)f(x)=f(y)+f(x/y)

f(x)=f(y)+f(x/y)

f(x/y)=f(x)-f(y)

2) If f(3)=1 and f(a) is greater than f(a-1)+2, find the range of the value of the real number a.

f(a)>f(a-1)+2

f(a)-f(a-1)>2

f(9)=f(3)+f(3)=2

f(a)-f(a-1)>f(9)

f(a/(a-1))>f(9)

f(x) increment function.

a/(a-1)>9

a-1>0 a>1

a>9a-9

a<8/9

No solution. a-1<0 a<1

a<9a-9

a>8/9

8/9

(1) Let m = 1 and n = 1.

f(mn)=f(1)=f(1)+f(1)

So f(1)=0

Let m=2, n=2 be substituted.

f(4)=f(2)+f(2)=1+1=2

2)f(a)+f(a-3)=f(a×(a-3))=f(a^2-3a)

f(x) is defined at (0, positive infinity).

a^2-3a>0

It is also an increment function.

To make f(a2-3a)>=2=f(4), then a2-3a>=4

Solution: a>4 or a<-1

If you don't understand, you can ask!!

Solution: 1. Let m=n=1 be established by f(mn)=f(m)+f(n) to obtain f(1)=f(1)+f(1).

So f(1)=0

Let m=n=2 be formed by f(mn)=f(m)+f(n) to obtain f(4)=f(2)+f(2).

f(2)=1 So f(4)=f(2)+f(2)=1+1=22, obtained from f(a)+f(a-3)<=2.

f(a)+f(a-3)<=f(4)

i.e. f(a(a-3))<=f(4).

Since the function f(x) is defined as an increasing function over (0, positive infinity), it is obtained from f(a(a-3))<=f(4).

a>0a-3>0

a(a-3)<=4

Solve this inequality group.

The value range of 3 with a is (3,4).

(1)f(2)=f(2*1)

Because f(mn) = f(m) + f(n) f(2) = f(1) + f(2) so f(1) = 0

f(4)=f(2*2)=f(2)+f(2)=2f(2)=2(2)f(a)+f(a-3)=f(a*(a-3))f(4)=2

Since f(x) is an increasing function.

So a 2-3 a<=4

So -1 "a" 4

Since f(x) is defined in the domain (0,+ a>=0 a-3>=0, 3"a"4