In the second year of junior high school, the person who asks for help will help

1) Set the price of x yuan. Then x must be less than 7 yuan per kilogram.

Then: 6 + [(7-x) The number of items that can be sold when the price is x yuan.

Then [6+2(7-x)]*x = total daily sales at the time of pricing x yuan = 50 yuan.

2) [6+2(7-x)]*xThis is the amount sold.

2*[6+2(7-x)] This is the cost price.

The two are subtracted even if the net profit is 24 yuan.

From this we get the equation. The second question is better to set up another meta. Solution: (1) Set the sales unit price to be x yuan, and the daily sales volume is 6 + (7-x) kg.

Column equation x*[6+(7-x) solves x=52) profit as total sales minus cost (i.e., purchase**).

Let the sales unit price be y yuan, and the column equation y*[6+(7-y) solves y=4 or y=8 from the problem, the sales unit price shall not be higher than 7 yuan, and shall not be less than 2 yuan, then y=4

Solution: (1) Set the sales unit price to be x yuan, and the daily sales volume is 6 + (7-x) kg.

Column equation x*[6+(7-x) solves x=52) profit as total sales minus cost (i.e., purchase**).

Let the sales unit price be y yuan, and the column equation y*[6+(7-y) solves y=4 or y=8 from the problem, the sales unit price shall not be higher than 7 yuan, and shall not be less than 2 yuan, then y=4

1) Set the unit price to be reduced by x yuan, then.

7－x）（6+x/

7－x)(6+2x)=50

42+14x－6x－2x2－50=0

x2－4x+4=0

x－2)2=0

x=2"x2" means x

When the unit price of the product is set at 5 yuan per kilogram, the total daily sales of the product is 50 yuan.

2) Let the unit price be reduced by Y yuan, then.

7－y）（6+y/ 2 （6+y/ )

5－y)(6+2y)=24

30+10y－6y－2y2－24=0

y2－2y－3=0

x－3)(x+1)=0

x=-1 does not hold x=3

When the sales unit price of the commodity is set at 4 yuan per kilogram, the total daily profit of the commodity is 24 yuan.

3x-m is less than or equal to 0

3x≤mx≤m/3

Positive integers are solved as 1

So 1 m 3 2

The value range of 3 m 6 m is 3 m 6

The solution of 3x-m<=0 is x<=m 3, and the solution of a positive integer is 1, then 1<=m 3<2

3<=m<6

m 3 cannot = 2, if m 3 = 2, then there is x< = 2, and the positive integer solution is 1 and 2.

Solution: 3x-m 0

x m 3 because the positive integer solution is 1

So 1 m 3 2

3≤m＜6

3x-m 0, 3x m, x m 3, the positive integer solution is only 1, then there is m 3<2, so the range of m is m<6

3x-m is less than or equal to 0

x=1, then 3-m is less than or equal to 0

3 is less than or equal to m

The value range of m is greater than or equal to 3

Constructing a Line Remember not to write a function once.

Put y to the left of the equation and the others to the right. Observe k and b

If k is equal and b is unequal, then parallel; If k is equal and b is equal, it coincides. If kb are unequal, they intersect. This determines the MN

The rest is solved by yourself (to add that if two straight lines are perpendicular, the slope k is negative to each other.) )

Convert the equation as:

2x-3y+m=0，3y=2x+m，y=2x/3+m/3n-1）x+6y-2=0，6y=（1-n）x+2，y=（1-n）/6+1/3

There is a unique solution when 2 3≠(1-n) 6, i.e., n≠-3.

When 2 3 = (1-n) 6 and m 3 = ≠1 3, i.e. n = -3, m ≠ 1 there is no solution.

When 2 3 = (1-n) 6 and m 3 = 1 3, i.e. n = -3 and m = 1 there are an infinite number of solutions.

Using the triangular median line parallel to the bottom edge, fh bc is obtained

ge//bc

Hence fh ge

The same goes for fg he ad

So the quadrilateral egfh is a parallelogram.

Happy studying.

In the triangle ABC, E is the midpoint of AB, G is the midpoint of AC, GC is the median, GC is parallel to BC, and in the same way HF is parallel to BC, so GC is parallel to HF.

Again, HE is parallel to GF

The quadrilateral row GEHF is a parallelogram (two sets of quadrilaterals with opposite sides parallel are parallelograms).

I don't have time, I'll have time to do it tomorrow Simple.

The question should add the condition, bac=90°, and the proof method can refer to huyudu.