Please help me solve a high school math problem, please help me solve a high school math problem?

Look at me, it's simple and straightforward.

a is the base, a>0, g(x)=2-ax is the subtraction function, (the base is greater than 0)f(x)=loga g(x) is the subtraction function, a>1, (the monotonicity of the composite function, that is, the internal and external functions increase in contrast, and the whole composite function is the subtraction function).

On [0,1] is the subtractive function, when x=1 2-ax is the smallest, at this time 2-a>0, a<2 (true number greater than 0).

In summary,1

Since y loga(2 ax) is a subtraction function of x on [0,1], a 0 and a ≠ 1 are known from the definition of the logarithmic function

1.When 0 < a < 1, satisfying 2 - ax is an increment function and obviously does not hold.

2.When a > 1, meet 2 - ax > 0 constant.

i.e. 2 - a > 0

So a < 2 gives 1 < a < 2

In summary: the value range of a is 1 < a < 2

A is greater than 1 when the function is subtractive and 2-ax is greater than 0

ax is greater than -2

Because x belongs to (

a greater than -2a less than 2a (

If you have any questions, please feel free to ask.

What about the topic? See if I will.

Questions. <>

When m=1, n=2, f(x)=ax m(1-x) n=ax(1-x)2=a(x 3-2x 2+x), so f'(x) = a(3x-1)(x-1), let f'（x）=0x=

x=1, i.e. the function is in x=1 3

of one-third square] and then minus one-second square, so that it is solved as 8 1000, s2:1 3x

s3: put-remove: original = 1000 8 to the 1 6th power (i.e. 125 to the 1 6th power).

Ah a, b, c three-point non-collinear vector ab equals vector bc uncollinear vector ab=vector bc=0 vector ab=(1,-4) vector bc=(x-3,2) solution gives x≠5 2(x r).

2) Because m is on a straight line oc, let 0m(6x,3x) ma=(2-6x,5-3x) mb=(3-6x, 1-3x) because ma mb

ma multiplied by mb=0 to get m as (2,1) or (22 5 ,11 5).

I did it last week Trust me Adopt it o( o thank you.

(1) The equation for finding the straight line ab is y=-4x+13, so that a, b, and c are not on the same straight line to form a triangle.

So, just remove the intersection of the line y=3 and the line ab y=3y=-4x+13

Solve the intersection point (,3).

Therefore, the value range of x is x≠

2) The linear oc equation y=x 2 is easily obtained

Let m(a,a 2).

Then the straight line ma is perpendicular to the straight line mb, and the following needs to be satisfied:

The product of the slopes of two straight lines is -1

k(ma)=(5-a/2)/(2-a)

k(mb)=(1-a/2)/(3-a)

5-a 2) (2-a)]*1-a 2) (3-a)]=-1 solution: a1=2 (rounded), a2=22 5

So m(22 5, 11 5).

(1) x is not equal to assume that the vector ab is equal to the vector bc, and it is enough to find that x is equal to).

2) m(2,1) or m(22 5,11 5).(Since m is on the straight line oc, let m(a,a 2) give the vector ma and the vector mb, so that the two vectors are multiplied by 0, and a is equal to 2 or 22 5, and then m is the above answer.) )

a^x+b^-b^x>=a^x^+2abx(1-x)+b^(1-x)^

(a + b -2ab) x + (2ab-2b -2a) x < = 0 )

(a-b) x - (a-b) x<=0 (a does not equal b) = x(x-1) < = 0

0<=x<=1

g'(x)=k/(x+1)

kx-1)/(x+1)²

k(x+1)-(kx-1)]/x+1)²

k+1)/(x+1)²

g(x) is the increasing function, g'(x)

So (k+1) (x+1).

Up (x+1) >0, so only k+1>0 is needed.

k>1f(x)=e g(x)0, so take the logarithm on both sides.

g(x)0).

t(0)<0,t(0)

1<0, satisfied.

t(x) is a non-increasing function, x>0).

t'(x)k+1)/(x+1)²-1/(x+1)k-x)/(x+1)²

t'(x) is a non-additive function when <=0.

So if for any x (0, + there is (k-x)<=0, there is only k<=0

Mathematical induction.

Let the left f(n) and the right g(n).

When n=1, left = ln(1+1*2) = ln(3)>0 right = 2*1-3

So on the left side and on the right side, the inequality holds.

Suppose n-1 is true (n>2).

f(n-1)>g(n-1)

then f(n)=f(n-1)+ln(1+n(n+1))g(n)=2*n-3=2*(n-1)-3+2=g(n-1)+2f(n)-g(n)=f(n-1)+ln(1+n(n+1))-g(n-1)+2).

f(n-1)-g(n-1)]+ln(1+n(n+1))-2ln(1+n(n+1))-2

Because n>2, ln(1+n(n+1)).

ln(7)>2

So f(n)-g(n)>0, f(n)>g(n), i.e., as long as f(n-1) >g(n-1), there is always f(n)>g(n), so the original inequality holds.