Help calculate a few junior high school math problems, junior high school math problems, how to calc

The first one of you is not typing?

2(m-n)^2+32

2[(m-n)^2-16]

2(m-n-4)(m-n+4)

16x^4y^4z^4-1

4x^2y^2z^2)^2-1

4x^2y^2z^2+1)(4x^2y^2z^2-1)

a+b)^3-a-b

a+b)^3-(a+b )

a+b)[(a+b)^2-1]

a+b)(a+b+1)(a+b-1)

4(x+y+z)^2-9(x+y+z)^2

2(x+y+z)]^2-[3(x+y+z)]^2

2(x+y+z)+3(x+y+z)][2(x+y+z)-3(x+y+z)]

5x+5y+5z）（-x-y-z）

80a^2(a+b)-45b^2(a+b)

a+b)(80a^2-45b^2)

a+b) (4 times root number 5 a + 3 times root number 5 b) (4 times root number 5 a-3 times root number 5 b).

I don't know if you have learned the root number, if you don't learn it, you can only decompose it to the second one].

2x-3y)^2-(3y-2x)^3

2x-3y)^2(2x-3y+1)

I don't know if you are in the second year of junior high school, we learned this in the second year of junior high school, I am now in the third year of junior high school, and I am sprinting for the high school entrance examination. In fact, this kind of problem is nothing more than to master [extracting common factors], [cross multiplication] and those formulas.

a 2-b 2=(a+b)(a-b)】【a+b) 2=(a 2+2ab+b 2)] and so on. You should also pay attention to symbols, first associate the formula when you see the question, and then see how to change the symbol into the form of which formula. But then you give a lot of points, and it's tiring to fight so much, but I used to come here often to ask questions, and no one was super upset when I asked, so I came to help you If you don't understand anything, you can also come and ask me.

In fact, my math score is not very good, and it is very important to be able to do factorization, because in the future, I often have to decompose the quadratic equation and quadratic functions.

I didn't understand what I was asking for.

x+x+x/2x/4+1=100

Solution: Multiply both sides of the equation by 4 to obtain.

4x+4x+2x+x+4=400

Move items, merge similar items, get.

11x=396

Divide both sides of the equation by 11.

x=36

You can just merge the same kind of terms, you can get 11 4x=99, then divide both sides by 11 to get 1, 4x=9, and finally get x=36.

Question 1: 210 1 20 252 (RMB) Question 2: Suppose that when the number of students is x, companies A and B charge the same amount.

Then: (x 1) 95 x 2 80 solution: x 13 3

Therefore, when x 4 x is a natural number, select Company A is cost-effective, and when x 4 x is a natural number, select Company B is cost-effective.

(1) Solution: If the price of this batch of goods is up to X yuan, then the price increase of each product is X-21 According to the title: X-210 20%X

Solution: x means that the price of this batch of goods cannot exceed yuan

2) The cost per person is x yuan, and there are y students, and the fee for company A according to the topic is: x+

The fee to be paid to Company B is: 80% (2+y)x

80%（2+y）x-（x+

A: If there are 6 students, the cost of the two companies is the same;

When the number of students is greater than 6, Company A saves money;

When the number of students is less than 6, Company B saves money

Solution: (1) 210x(1+20%)=252 yuan from the meaning of the question: the maximum can not exceed 252 yuan.

2) [I don't want to do it] This question should be discussed in three situations - A is cost-effective, B is cost-effective and equal with X students, and the ticket price is A (A is not just a decoration) is known by the meaning of the question:

x＜13/3

x=13/3

x＞13/3

x is a positive integer.

x 4 or x 5

Therefore, when the number of students does not exceed 4, it is cost-effective.

Not less than 5 people B is cost-effective.

1、b2、a>b

3, y=, 4, from the graph can be seen that y and x are a one-time function equation, that is, y=kx+b, the value on the graph can be substituted into the formula is y=-5x+30

1) When y=0, x=6, that is, a tank of oil can be used for the car for 6 hours (2) when x=1, y=25, that is, 30-25=5, so the car consumes 5 liters of oil for 1 hour.

3) When y<5, that is, -5x+30<5, that is, x>5, so when the car driving time is less than 5 hours, the car will automatically alarm.

1. b

2.Bring -5 in to get the value of a, and 4 in to get the value of b) 3 ⑴y=

4.According to two points (0 30）（5. 0）.The primary function y= -5x+30 When y=0, x=6 A tank of fuel can be used for the car for 6 hours.

When x=1, y=25

The amount of oil remaining in the tank is less than 5 liters, i.e. y 5,( 5x+30 5 ) solution x 5

1、b2、＞

3、（1）y=

2) y = 4, (1) a tank of oil can be used for 6 hours of driving.

2) The car consumes 1 = 30 liters of fuel for every 5 hours driven.

3) (30-5) Automatic alarm after 5=5 hours.

You can set a one-dimensional one-time function.

y=ax+b

Substituting (0,30) and (6,0) gives y=-5x+30

1、b2、＞

3、（1）y=

2) x 1 = 30 6 solution x = 5 i.e. fuel consumption of 5 liters per hour (3) 30-5 = 25 liters.

25 x=30 6 solution x=5 that is, driving for 5 hours to start the alarm.

1 hail a b 0 (when a=-2, b=-1) the value of a+b is still negative (-3).

The beam value of b-a is the positive branch movement (1).

The values of a-b are still negative (-1).

The values of a-b are positive (3).

It's much easier to assign a value to a problem like this! ）

It's not interesting if the homework is not done by Hu Lu himself. If you ask questions on the Internet, it's better to search for them tomorrow morning, and go to school early to find some top students who are good at doing the calendar. Or let them teach you. It's a bit of a hassle to answer these questions online. You'd better go to bed early.

3.(1) a2 = 2 a3 = -1 a4 = one-half.

2) a2008-a2009 = minus three-two.

Question 1: First find x and y and then find their sum x=(a-b) (c-a).

y=(b-c) (c-a) their sum is -1

Question 2: First of all, we must know the maximum and minimum number of intersections, and the most should be found like this, for example, two intersections are one intersection, two intersections are added when the third is added, three intersections are added when the fourth is added, and so on, that is, 1+2+3+4+5+6 is the maximum total number of intersections, 21, the least is 1, and the sum of them is 22

Question 3: The letter to be emphasized is transformed into a known quantitative relation a-b=5 and c-b=10, and the proposed common factor after c-a=5 is transformed into a(a-b)+b(b-c)+c(c-a)=5a-10b-5c=5a-5b-5b+5c=5(a-b)+5(c-b)=75

Question 4: You can't find a specific value in this question, but you can know that it must be a negative number, and after the general score, it is [-(a-b)(b-c)-(a-c)(a-c)] (a-b)(b-c)(c-a) is a negative number.

Let x a-b=y b-c=1 c-a=k, then x+y a-b+b-c=x+y a-c=k is obtained according to the properties of the ratio

1／c-a=k，∴x+y／a-c=1／c-a∴x+y=-1

Let the square of a + the square of b + the square of c - ab-bc-ca=k, and then multiply both sides by 2 at the same time, and then make (c-a) square + (b-a) square + (b-c) square = 2k

According to the known a-b=5 and c-b=10, and then calculate c-a=5 by substituting it, we can calculate that it is equal to 75

There is something to deal with, these kinds of questions are Olympiad questions, if you have a good foundation, you can do it appropriately.

1. Is x a-b (x a)-b or x (a-b)?!

2.Intersection of Straight Lines: Maximum (n(n-1)) 2 and minimum 1 (n=7) a+b=22

3.The original formula = (a-b) 2+(b-c) 2+(a-c) 2 2 is given by the known a-b=5, c-b=10, then a-c=-5 is substituted into the above equation to get 754Please make the title clear...

If any two of the 7 straight lines in the plane intersect, and the intersection points have a maximum and a at least, then a+b=?

Answer: The intersection of a straight line is 0

2 0+13 0+1+24 0+1+2+3 The intersection of 7 straight lines is 0+1+2+3+4+5+6=21 Knowing a-b=5 and c-b=10, then the square of a + the square of b + the square of c - ab-bc-ca is equal to?

Answer: a-b=5 and c-b=10 are subtracted to get a-c=-5 and the three formulas are squared and added to give 2 (a2+b2+c2+2ab+2ac+2ba)=150, so the result is equal to 75

x=(6+-root(36-4k)) 2=3+-root(9-k)) is substituted into the following formula.

3 + root number (9-k)) 2 (3-root number (9-k)) 2-3-root number (9-k) -3 + root number (9-k).

9-(9-k)]^2-6

k^2-6=115

k^2=121

k=+-11

deta=36-4k because there are two real roots, so deta>0 so k=-11

1) By Vinda's theorem, x1x2=k, x1+x2=6. and from x1 x2 -x1-x2=k -6=115.

k²=121，k=±11；

2) x1 +x2 +8=(x1+x2) -2x1x2+8=36 22+8=22 or 66

1) Because x1x2=c a=k, x1+x2=-b a=6, then x1 x2 -x1-x2=115

Obtain: the square of (x1x2)--(x1+x2)=115, that is: the square of (x1x2)-6=115, so the square of (x1x2)=121, so k=plus or minus 11, when k=11, the square of (-6) -4*1*11 is less than 0, so 11 is not true, so k=-11

2) The square of x1 +x2 +8=(x1+x2) -- the square of 2x1x2+8=6 - 2*(-11) + 8=66

1. Solution: It takes x and y days for A and B to build this road respectively.

1 x) + (1 y) = 1 6 is , the road that A repairs every day + the road that B repairs every day = the road that A and B repair every day.

40%*x-(13 and 1 3)%*y=2 is, the number of days of 40% of the road of A - the number of days of the road of B repair (13 and 1 3)% = 2 days and then solve the system of equations.

2. Solution: A rides back to pick up C, and meets at D place, it can be seen that (AC+DC) vehicle speed = AD walking speed Distance divided by speed = time, and 3 times the relationship between speed and speed, AC+DC=3AD is the midpoint of AC.

In the same way, we can find that c is the midpoint of db, that is, we can find the distance between acs.

3、（5a+2）（a-4）=0

The second one uses the root finding formula.

I hope the landlord is satisfied, and the landlord is welcome to ask for advice and discuss math problems together.