Let a be a rational number, x 1, and verify that if 0 a 1, then 1 x a 1 ax

Let f(x)=(1+x) a-(1+ax), then f'(x)=a{(1+x) (a-1)]-a, extracts the common factor a, so f'(x)=a[(1+x)^(a-1)-1]

Because of 00, f(x) is an increasing function on [-1,0], and because f(0)=0, f(x) < 0 when x belongs to [-1,0].

When 1+x>1, i.e. x>0, there is f'(x) < 0, so f(x) is at (0, +% zero to positive infinity.

is a subtraction function, and because f(0)=0, the monotonicity of x>0 is determined by the monotonicity, and f(x) 0 is always f(x) 0, that is, the proposition is proved.

As for the second question, the derivative can also be used to prove it, so I won't say more.

I think a handwritten version ......Let's make it up or see.

Thanks to rchlch for the ideas.

There is a way to prove the problem, which is to deduce it backwards, for example, in this problem, let us prove that is: (x) (x+a)>(y) (y+b), we take the reciprocal at both ends of the formula, we can get, a x hehe, will you?

special a=1 rounded;

When a<1, for x>0, (agenus 1) x 1<0 is constant.

x>1x-1>0

ax(x-1)+x>4(x-1)

ax^2-(a+3)x+4>0

A> Number Brigade 0 and Potato Stool Old Mess (A+3) 2-16a

Question 1. Imagery.

Draw the | in the same coordinate systemx+1|AND |x-2|In the image, it can be intuitively seen that f(x) has a minimum value when x=1 2.

To be more specific, the function is equal to 2-x when x<=, and the function is equal to x+1 if it is greater than the time width. These can be seen from the pictures.

Question 2. If |x-a|<εx-c|< Proof.

a-c|<2ε

x-a|<ε

xεx-c|<ε

x <-c< x Eq. 2

Equation 1 and Booth 2 are added.

Get. -2 plus is definitely worth it.

a-c|<2ε

Four positive whole shed cultivation number.

So these four positive integers are 1, 2, 3, 4

Therefore, drawing the number line can know that the value range of a is 5, and it is not easy to type a 4.

Hope is bright and quiet, and wide hee-hee.

||x-a|-b|=3

x-a|=3+b

Or. x-a|=b-3

If b+3, b-3 is non-negative.

And if one of them is 0

then you get 3 solutions.

If none of them are zero.

then there are 4 solutions.

Therefore b = 3

Solution: x 0, x+1 x

2 (equal sign if and only if x = 1), x x 2

3x+11/(x+1/x

i.e. x (x 2

3x+1) has a maximum value of 1 5

So the answer is a 1 5