The solution of math problems in the first year of junior high school requires a process and urgency

1. Solution: Original formula = (x+2)(x-2)(4+x )-2+x ) =(x -4)(x +4)-(2+x) =-4x -20

Substituting x=-3 2 yields: -4x -20=-4(-3 2) -20=-29

2. Solution equation: (2x-1) -4(x+2)(x-2)=x-1 Solution: 4 x -4x+1-4(x -4)=x-15x=-18

So, x=

3. Known(a+b) =24, (a-b) =20, find the value of ab.

Solution: a +2ab+b = 24---1).

a²-2ab+b²=20---2)

1)-(2) gets: 4ab=4

So, ab=1

Simplify, then evaluate: (x+2)(x-2)(4+x)-2+x) where x=-three-two.

x²-4））（4+x²)-2+x²)²x∧4-16-(2+x²)²x∧4-16-4-4x²-x∧4

20-4x²

Solve the equation: (2x-1) -4(x+2)(x-2)=x-14x-4x+1-4x +16-x+1=05x=-18

x=18/5

Known(a+b) =24, a-b) =20, find the value of ab.

4ab=4 ab=1

Suppose x cm lead is used, then the volume of 10 lead balls is:

s=10*(4/3)*pi*(12/2)^3;

Since the volumes should be equal, so.

s=x*pi*(12/2)^2;

The above two formulas are equal:

x*pi*(12 2) 2=10*(4 3)*pi(12 2) 3 can find that x is equal to 80 centimeters.

1)40

2) Set your birthday date to x

Then the upper number is x-7, the lower number is x+7, the left number is x-1, and the right number is x+1

According to the title, 4x=60

x=153)x+y=105

8x+12y=1500-418=1082

Solution x=y=unreasonable 4) Suppose you want to buy x books, and the unit price is Y yuan.

According to the title: xy=4 5xy+20

The solution is xy=100

That is to say, he spent the same amount of money to buy a total of 100 yuan of books, less than 100 is not worth membership, and more than 100 is worth membership.

5) Set the width x to the length of 4x

According to the title: x+4x=5x=100

x = 20 and the area is 20 * 80 = 1600

6) Set the list price x

3/5x-58=32

x=150

Since the answer is 0, the terms of this polynomial are homogeneous, i.e., *y to the power of x, -3xy to the power, and *xy to the same kind.

So, =1, =2, =1-3)=2.

If the number of gymnastics teams is 5x, the number of basketball teams is 6x, and the number of volleyball teams is 10x-5

6x + 3*(5x) = 42

21x = 42

x = 2 gymnastics team size = 5 * 2 = 10 (people) basketball team number = 6 * 2 = 12 (people).

Number of volleyball team = 10*2-5=15 (people).

If the gymnastics team is x, the basketball team will be 6 5x, and the volleyball team will be 2x+5

6/5x+3x=42

x=10

There are 5t people in gymnastics, 6t in basketball, and 2x5t-5 in volleyball.

The columnable equation is 6t+5tx3=42

There are 10 gymnastics, 12 basketball, and 15 volleyball.

Solution: Suppose you pay 5 RMB x and 2 RMB y, and list the equation: 5x+2y=19There are two integer solutions to the eating equation: 1:x=3, y=:x=1, and y=7. Hope it helps.

There are x sheets for $5 and y sheets for $2.

then 5x+2y=19

There is the first one. 1 ticket for 5 yuan, 7 tickets for 2 yuan.

The second. 3 for 5 yuan, 2 for 2 yuan.

I'll answer.

The vendor gave the farmer a total of buckets and rice.

The weight of the iron drum is not more than 19 catties, assuming that the weight of the iron drum is 1

Corn 45 iron drum 1 rice 18

Iron bucket 1 rice.

Let's say the weight of the iron bucket is 2

Corn 44 iron drum 2 rice.

Iron bucket 2 rice.

Finally, it is assumed that the weight of the iron drum is 18

Corn 28 iron buckets 18 rice.

Iron bucket 18 rice.

Therefore, the peasant woman was not profitable in this transaction, and the greater the weight of the bucket, the greater her loss.

If you don't understand, you can contact me, I'm helping you answer.

Set the barrel weight x.

then the corn weighs 46 - x

It should be replaced with rice (46 - x).

Rice - x

Difference = (46 - x) - x) = - 2x 5 - x = 3x 5 > 0

Therefore, it is not cost-effective to exchange less rice than should be exchanged.

Not cost-effective!

Assuming that the weight of the barrel is 6 catties, the weight of the corn is 40 catties.

The fair case should be: 40 catties of rice.

But in fact, it was a catty, and the peasant woman only got a catty of rice, which was less than the fair situation.

If it is not cost-effective, assuming that the barrel weighs 1 catty, the original corn is 46-1=45 catties, which should be able to exchange for 45 catties of rice, but she actually only changed catties.

Solution: Let the flow rate of the first pump be a, the flow rate of the second pump is b, the first stage two pumps at the same time to the pool water injection time t1, (a+b)*t1=2 9-1 18 3 18, the first pumping of the second stage for 81 minutes flow rate 81a=b*t1, the second pumping of the third stage for 49 minutes flow rate 49b a*t1, the sum of the pumping volume of the second stage and the third stage is equal to the first stage is also 3 18, by: a b=7 9, t1=63 minutes, and capacity:

1-1 18-3 18-3 18=11 18, the pool needs to be filled with water and need to be filled with water together x minutes: 63:3 18=x:

11 18, x = 231 minutes.

The two pumps need to be filled together for 231 minutes before the pool can be filled.