Super difficult math problems in the second year of junior high school, the most difficult math prob

x+y+z=30

3x+y-z=50

5x+4y+2z=m

Take m as a constant and solve this system of equations.

x=140-m

y=-240+2m

z=130-m

Because x, y, z, are all non-negative numbers.

140-m>=0

240+2m>=0

130-m>=0

So m<=140, m>=120, m<=130, so 120<=m<=130

It's late again

The ground floor is doing the right thing.

From the first two calculations, we can get that y = 40 - 2x; z = x - 10 then: m = 5x + 4*(40 - 2x) + 2*(x - 10).

5x + 160 - 8x + 2x - 20140 - x.

The rest is fine, hehe

By Equation 1*3, subtract Equation 2We get y+2z=20, y=20-2z, and equation 2 subtracts equation 1 to get it, 2x-2z=20, x=10+zBring in m = 50 + 5 z + 80 - 8 z + 2 z = 130-z....

Now we have the condition that "=0,>=0,>=0 to find the range of z values 0<=z<=10, bring in m=130-z to get 120<=m<=130... You do the math. I may have miscalculated ...

Here's how it works.

I just saw the first floor. He's right!

1.Container A has 30 liters of 15% brine and container B has 20 liters of 18% brine, if the same amount of water is added to each of the two containers so that their concentration is equal, then how much water is added?

2.If A does a project alone, it will be completed at the specified time; If B does it alone, it will take 3 more days than the specified date to complete, and now after 2 days of cooperation between teams A and B, team B will do it alone, and it will be completed on the specified date.

3.A pool of water has a water.

B has two water inlet pipes and opens A at the same time.

After 4 hours of pipe B, pipe B was closed, and pipe A used another 6 hours to fill the pool. The water injection of tube A is the same for 2 hours and 30 minutes and tube B for 2 hours. How many hours does it take to open the two pipes of A and B separately to fill the empty water pool?

4.Last winter, Manager Wang used 80,000 yuan to buy a batch of clothing, a 58 yuan per piece of sales, the result is not to seek, and then with 17,600 yuan enough to enter the quantity is twice the first time, the unit price is more expensive than the first time 4 yuan of the same clothing continues to sell, considering the change of seasons, and finally the remaining 150 pieces are sold at a 20% discount, and they are sold out soon. Q: How many costumes are in this batch?

Is the clothing store's business a profit or a loss? Why?

Solution. 1. X liters of water should be added.

1) Find the salt content of container A and B.

2) The concentration is equal after each addition of x liters of water.

30*15%/（30+x）=20*18%/(20+x)2、

Set the specified time x and consider the total workload as 1. Then A does 1 x work every day, and B does it every day.

1/(x+3)

The work completed by A and B in 2 days is:

2*1/x2*1/(x+3)

The remaining work after 2 days of cooperation between A and B.

1-[2*1/x

2*1/（x+3）]

The number of days it takes for B to do the remaining work alone.

Remaining Daily Workload B's workload per day.

1-[2*1/x

2*1/（x+3）]]/[1/(x+3)]

The specified number of days = the number of days of cooperation between A and B.

The number of days it takes for B to do the rest of the work alone.

i.e. x=2+1-[2*1 x

2*1/（x+3）]]/[1/(x+3)]

3. The total water volume of the pool is 1, A needs x hours to fill the pool, and B needs Y hours. Then A injects 1 x of water per hour, and B injects 1 y. per hour

4/x+4/y

6 x1 Equation 1 (while turning on A.

After 4 hours, B closed B, and A took another 6 hours to fill the pool).

2*1 Y equation 2 (the water injection of tube A is the same for 2 hours and 30 minutes and tube B for 2 hours) 4

If x pieces are purchased for the first time, 2x pieces will be purchased for the second time.

Sales for the first time.

58x sales for the second time.

58+4）*（2x-150)

I can't do the 4th question.,Is there a typo?。。。

1.Two trains, A and B, leave from A and B at a constant speed at the same time, car A goes to city B, and car B goes to city A Due to the ink covering, the figure provides only a part of the function image of the distance between the two cars from city B S A (km), S B (km) and travel time t (hour) (1) The speed of car B is km-h;

2) Find the functional relations of S A, S B and T respectively (no need to write the range of values of t);

3) Find the distance between the two cities, and why the two cars meet;

4) When two cars are 300 kilometers apart, find the value of t

I think this question is more difficult, hehe.

Whether you can do it or not is not the same as "the most difficult math problem to do in the second year of junior high school". Further, there is no "most difficult math problem to do in the second year of junior high school". Definitely not, some just said the wrong thing.

Recently, it has been strict, and it is difficult to find a child in Yizhou

The garbage is dead, and I can't do it (in fact, I won't either).

Whether you can do it or not is not the same as "the most difficult math problem to do in the second year of junior high school". Further, there is no "most difficult math problem to do in the second year of junior high school". Definitely not, some just said the wrong thing.

Asked the right person, Wang cheap baby

Solution: (1) Method: Do a along the corresponding point of the straight line 0m A1 and do a along the corresponding point of the straight line 0n A2

Connect A1 A2, the intersection of A1 and OM is B, and the intersection of A2 and ON is C.

Analysis: ab in abc at this time, ac is equal to a a1, a a2. So the circumference of the ABC is minimal.

Reason: The shortest line segment between two points.

2) Method: Make a vertical line of straight L through B.

Intersect the straight line l at the point c

Connection AC analysis: Since the absolute value of the CA shear CB is maximized, the CB should be the shortest.

Rationale: The shortest distance from a point to a line is its perpendicular segment.

3) Method: 1 is equal to 2, ab ac, db dc

pdc pdb

bp pc gets the inequality: ab + bp ac + cp

Solve the inequality to get:

ab+bp＞ac+cp）

ab-bp＞ac-cp）

ab-ac＞bp-cp

Analysis: Because 1 is equal to 2, ab ac, and only db dc ab ac so db dc. Then, look at the image to see the PDC PDB.

So say bp pc. Finally, AB AC and BP PC are combined to obtain the inequality group AB+BP AC+CP, and the junction inequality is obtained: PB - PC AB - AC

Solution: (1) Straight line L parallel to BC

Then BCE= OEC, DCF= OFC CE divides BCA

bce=∠oce

oec=∠oce

OE = OCCF divides the DCA

dcf=∠ocf

ofc=∠ocf

of=ocoe=of

When O is the midpoint of AC, the quadrilateral AECF is rectangular.

oa=oc and oe=of=oc

The quadrilateral AECF is a parallelogram.

bce=∠oce，∠dcf=∠ocf

ecf=∠oce+∠ocf=90°

The quadrilateral AECF is rectangular.

Sorry to do the task

Solution: (1) Straight line L parallel to BC

Then BCE= OEC, DCF= OFC CE divides BCA

bce=∠oce

oec=∠oce

OE = OCCF divides the DCA

dcf=∠ocf

ofc=∠ocf

of=ocoe=of

When O is the midpoint of AC, the quadrilateral AECF is rectangular.

oa=oc and oe=of=oc

The quadrilateral AECF is a parallelogram.

bce=∠oce，∠dcf=∠ocf

ecf=∠oce+∠ocf=90°

The quadrilateral AECF is rectangular.

From the title, it can be obtained: y1=k1(x+1).

y2=k2 x k1, k2 is not 0

y=y1+y2=k1（x+1）+k2/x

put x=1, y=0; x=4 and y=9 are substituted into 0=k1*(1+1)+k2 1=2k1+k29=5k1+k2 4, respectively

k1=2 k2=-4

y=2(x+1)-4/x

y=y1+y2

Let y1=a(x+1) y2=b x

i.e. y=a(x+1)+b x

Bring y=0 when x=1 and y=9 when x=4 into the above equation respectively:

2a+b=0 5a+b/4=0

Solving this system of binary equations gives a=2, b=-4, i.e., y=2x-4 x+2

Let y1=a(x+1); y2=b x,y=y1+y2=a(x+1)+b x, substituting x=1,y=0 and x=4,y=9 to get 2a+b=0,5a+b 4=9,the binary system of linear equations is solved:a=2,b=-4,the functional relationship between y and x is:

y=2(x+1)-4/x=2x-4/x+2

Assumption: y1=k1 (x+1).

y2=k2×x

From x=1,y=0: 2 k1+k2=0

From x=4,y=9 we get: 5 k1+4 k2=9 from the above two equations: k1 = 3 k2=-6

After bringing k1 and k2 into the hypothetical formula, we can bring y=y1+y2 to get y=-3x+3

Solution: Because y1 is proportional to x+1, let y1=k1(x+1), and y2 is inversely proportional to x, so let y2=k2 x (k1, k2 is not 0) so: y=y1+y2=k1(x+1)+k2 x put x=1, y=0; x=4 and y=9 are respectively substituted into the above equation to obtain:

0=2k1+k2

9=5k1+k2/4

Solve this equation to obtain: k1=2 k2=-4

So: y=2(x+1)-4 x