Senior 2 Mathematics Question 4 Ask for a detailed solution process

Move to the left with a 2-2a+1>0

a-1) 2>0, because a>1, the inequality is constant.

The left side a*3+1=(a+1)(a 2-a+1), the right a 2+a=a(a+1), move the term to the left.

a+1)(a 2-2a+1)=(a+1)(a-1) 2, because a>1, the inequality is constant.

Moving to the left has a 3(a-1) + a(a-1) = a(a-1) (a 2+1), and since a > 1, the inequality is constant.

Because a>1, it is sufficient to examine whether the inequality A 4+1>A 3+A 2 holds.

Moving to the left has a 3(a-1)-(a+1)(a-1) = (a-1) (a 3-a-1).

Take a = have a 3-a-1<0, the inequality does not hold.

In summary, the answer is 1,2,3

This kind of question is generally simpler with the assignment method. Algebra a minimum integer 2, as long as one does not conform to the equation, it is not constant. However, in order to insure more generations of 10, I guess it's okay to have a lot in my heart.

a-1) =a -2a+1>0 then a +1>2a (1) is correct.

a -a +1-a = a (a-1)-(a-1) = (a-1)(a -1)> 0 a +1>a +a (2) correct.

a) a +a -a=a ( a-1)+a(a-1)=(a-1)(a +a)>0 (3) Correct.

a^5-a^4+a-a³=a^4(a-1)+a(1-a²)=(a-1)(a^4-a-1)=(a-1)[(a²+1)(a²-1)-a)

a-1)(a +1-a)=(a-1)[a(a-1)+1]>0 When a>2, correct, when 1 chooses (1)(2)(3).

Knowing that f(x+8) is an even function, then f(x) is an even function after translating 8 units to the left, according to the nature of the even function, let g(x) = f(x+8), then g(x) is an even function, then.

g(-x)=f(-x+8)

Because g(-x) = g(x).

So f(8-x)=f(8+x).

Therefore, the axis of symmetry of f(x) is x=8, because f(x) is a subtraction function at x 8, so x 8 is an increasing function, and the function at f(8) has a maximum value, and the farther away from 8 is the value of the function that compares the size of the two sides, the smaller the value of the function, so that it is easy to compare.

Monotonically incremental. Let x2 x1 0, then x2 0, x1 0, x2-x1 0.

f(x2)-f(x1) x2- x1 ( x2+ x1) (x2-x1) 0, i.e. f(x2) f(x1).

So f(x) is an increment on [0,+.

<> can't see the trembling finch understands, you can ask me about the cave things in the morning.

If there are x stations, then each station must have a ticket to any other station, that is, each station must have (x-1) kind of ticket. Huai state file.

So, x(x-1) = 132

x²-x-132=0

x-12)(x+11)=0

x = 12 or -11 (rounded).

So, 132 kinds. Repent.

<> ginseng trapped things, Huai Kao, cover the ant thoughts.

I only saw the first question ......f(

Since the period of f(x) Zhouqing is 4, because the source of this x [6,7] also satisfies f(x) =x, so f( =

I don't think I have enough words.

Are there two words that are not illuminated?

The latter one is not used, and it is estimated that even functions and periods can be used.

Solution: Let the equation for the circumscribed circle of the triangle AOB be (x-a) 2+(y-b) 2=r 2

From the meaning of the title, (4-a) 2+b 2=r 2

a^2+(3-b)^2=r^2

a^2+b^2=r^2

The solution is composed of the above three equations, and the system of equations is a=2, b=3, 2, r=19 4, so the equation for the triangle AOB circumscribed circle is (x-2) 2+(y-3 2) 2=19 4