Three math problems in the first year of junior high school, master solutions!

1.The coefficient of x 2 and x without the primary quadratic term is 0.

2m - 1 = 0

2-3n=0

So m = 1 2, n = 2 3

2.There is already a 3y 5, as long as the 2x 3y n is not more than 5 times.

i.e. n < = 2and n is a positive integer n, so n=1 or n=2

3. (a-3)x^2y^|a| )b + 2) ?

5. The coefficient of the term is not 0, so that a-3 is not equal to 0, that is, a is not equal to 3

To the power of 5 you get 2 +|a|= 5, i.e. a= plus or minus 3

And a is not equal to 3, so a = - 3

The monomial is (b + 2) = 0i.e. b = -2

Bring in A2 - 3AB + B2 = (A - B) 2 - AB = -5

1.Since there is no quadratic term, so 2m-1=0 m=1 2 Because there is no quadratic term, so 2-3n=0 so n=2 32 Because the equation is a fifth-order polynomial, so 3+n 5 n 2 and n is a positive integer So n=1 or 2

Question 3 is a bit confusing.

1) 2m-1=o

2-3n=0

m=,n=2/3

2) If your second term is 2x times 3y to the nth power, then n=1,2,3,4 if 2x to the 3rd power multiplied by y to the nth power, then n=1,23) a +2=5, b+2=0, a-3 is not equal to 0, so a=-3, b=-2

Original = -5

Why can't I be in the fourth year of junior high school?

The difficulty is relatively large, in which the answer to the first question is 1, the second question cannot be solved, and the third question is obviously a binary quadratic equation, which cannot be solved, which should be high school.

Dadizhiling, hello:

Solution: (1) Set the purchase price of M-shaped clothing to be X yuan.

The price of the sale is 75 yuan, and the sales can still make a profit of 50% according to the 20% discount, then the equation can be listed as:

X 40 means that the purchase price of M-shaped clothing is 40 yuan.

2) The price is 75 yuan at the time of sale, and the price of each piece is reduced by 20% on the basis of 20% off.

The actual selling price of the M-shaped clothing to carry out the ** activity is 75

The profit from sales is 60 x 40 20 x.

And the functional relationship between the number of sales per day y (pieces) and the price reduction x (yuan) is y 20 4x The profit obtained from the daily sale of m-shaped clothing during the ** period:

w＝（20－x）（20＋4x）

4x^2＋60x＋400

4（x－15/2）^2＋625．

When x 15 2 yuan), the maximum profit w is 625 yuan.

(1) The price of the sale is 75 yuan, and the profit can be made by selling at 20% off (75 * 80% - purchase price) = purchase price * 50%.

Solution purchase price = 40 yuan.

2) w = quantity sold per day y (pieces) * how much money earned per piece = y * 40 * 50% -x).

20+4x) *20-x)

4x^2+60x+400

4(x^2-15x-100)

4[(x-15/2)^2 -225/4 -100]=-4(x-15/2)^2+625

When x=15 2=yuan, w is up to 625 yuan.

Solution: If the uphill road distance is x, the flat road distance is y, and the downhill road distance is z, then [(x+z) 150]+(2y 200)+[x+z) 240]=31

x+y+z=3000 ②

The equation is deformed to x+z=(37200-12y) 13 and the equation is brought into the equation y+(37200-12y) 13=3000

13y+37200-12y=39000

The solution is y=1800

So the flat road is 1800m

It can be seen that the time taken to walk a round trip uphill and downhill is 31-(2y 200)=31-18=13min

To rush back to the flat road in time, the maximum time to complete the blind meditation is 35-7-13 = 15min, the minimum speed is (1800*2) 15 = 240m min, so the minimum should be increased by 240-200 = 40m minA: ......

Solution:1Let the concentration range of this disinfectant be x

Then there are: 20%<[30 15%+50x) (30+50)]<35%: 23%2Let the cost of A and B be X and Y yuan respectively, then there is:

x+y=600;

Solution: x=425, y=175

3.Yes; Because the mineral water bought in class B is 6% off, and the mineral water in class A is obviously not;

If you buy x bottles on the first day of class A and Y bottles on the second day, you have: (because the second day's one is greater than the first day).

1* x+y=70;（302* x+y=70;(202x+;

Solution: x=45;y=25 (not satisfied).

3* x+y=70;(x<20)

2x+;Solution: x=; y=;(Not satisfied).

Therefore: as long as the mineral water bought on the first day is between 30 and 35.