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A typical ternary inequality application problem.

As the title knows, the number of first, second, and third grades is piece, order.

The unit price of the first, second and third class products is X million yuan, Y million yuan and Z million yuan respectively, and the profit model without loss is transformed into a mathematical formula: 20X + 30Y + 50Z 500, which is simplified to 2X + 3Y + 5Z 50

Assuming that the second-class product is sold according to the average price, that is, y=5, then the original formula is simplified to 2x+5z 35, when x=5, z=3, no loss;

When x=6, z=, no loss;

When x=7, z=, no loss;

When x=8, z=, no loss;

When x=9, z=, no loss;

When x=10, z=3, no loss;

Similarly, let y be different under different x values, and then flexibly adjust according to actual needs.

The price of first, second and third class products can be used.

Thanks, hope.

First-class unit price: 50000 (100*20%)=2500 yuan.

The unit price of second-class products: 50000 (100*30%) is about 1667 yuan.

Unit price of third-class products: 50000 (100*50%)=1000 yuan.

If the first class is X, then the second class is X-10000, the third and fourth class are X-20000, and the fifth class is X-25000

Then 13500 * [45% * x + 25% * (x-10000) + (15% + 10) * (x-20000) + 5% * (x-25000)] = 432000000, that is, 45% * x + 25% * (x-10000) + (15% + 10%) * x - 20000) + 5% * (x-25000) = 32000

The solution is x=11950

Hello lz, this is an indefinite equation, in the absence of other factors, x, y, z solutions have an infinite number of x, y, z can only be used one of x, y, z to represent the other 2 numbers, so the abc goods are x, y, z yuan 3x, 7y, z=315---1) 4x+10y+z=420---2) x>0, y>0, z>Qi with paragraph 0 (2)-(1), get x+3y=105x=105-3y---3) substitute (3) into (1)3(105-3y)+7y+z=315z=2y, today's x+y+z=105-3y+y+y+2y=105, so this person should pay 105 yuan.

Note: Even if the result of x+y+z is related to y and is not exactly eliminated, the result can also be regarded as a high-reputation function about y, in this problem the definition field of y is 0 If the result of x+2y+z is asked, then the result is 105+y, so the conclusion is that this person should prepare at least 140 yuan, and the money he paid is a number of 105 140.

If the first-class goods ** are empty x, then the second-class is X-10000, the third-class repentance, the fourth-class is X-20000, and the fifth-class is X-25000

Then 13500 * [45% * x + 25% * (x-10000) + (15% + 10) * (x-20000) late blind + 5% * (x-25000)] = that is, 45% * x + 25% * (x-10000) + (15% + 10%) * x - 20000) + 5% * (x-25000) = 32000

The solution is x=11950

1): The analytic formula of l y=-x+4

2): 43): t=or 1

If you don't understand something, ask again.

1) Directly substitute the function f(x)=x2+x into the inequality and simplify the solution

2) First, substitute the function f(x)=x2+x into the equation f(ax)-ax+1=5(a 1), and the equation f(ax)-ax+1=5(a 1) has a solution in c, and convert it into ax has a solution in a certain range

3) Find a first and then b, use a b to transform into an inequality group, and the solution can be done

Answer: Solution: (1) The original inequality can be converted to 2x2 2|x|, when x 0, 2x2 2x, the solution is 0 x 1 (2 points).

When x 0, 2x2 -2x, the solution is -1 x 0, so c = [-1, 1] (4 points).

2) (ax)2-(a-1)ax-5=0 from f(ax)-ax+1-5=0

Let ax=u, because x [-1,1], so u [1a,a].

Then the problem is transformed into finding u2 (a 1) u 5 0 with a solution (6 points) in [1a,a].

7 points) from the existence theorem of images and roots.

h(1a) 1a2 1+1a 5 0h(a) a2 (a 1)a 5 0 (9 points).

A 5 (10 points) is answered

3) a [ 14,2]g (x) = 3x2-3t 0 (because t 0).

So g(x) x3 3tx+t2, monotonically increasing on x [0,1].

So the range of the function g(x) is b[t2,1 52t] (13 points).

Because a b, so.

T2 142 1 52t solution t Natuan 12 (16 points).

The sum of the first n terms of the equal difference series is: sn=na1+n(n-1)d 2 or sn=n(a1+an) 2

Bring it in and get sn=-10n+n(n-1)=n 2-11n

Question 17: x=3 sides, y = 9 sides are bright and thick.

Question 18: 7+9> hit key empty x and |x-7|<9 and |x-9|<7 gets 2< x "Blind 16< P>