Junior high school math problems, junior high school math problems

exists, shifts the term to obtain: -m-2>(3-m)x, and it is easy to know that if m exists, the system of equations:

3-m>0 one.

m-2) (3-m)=-4 two formulas.

Must be established at the same time.

The solution of the second formula gives m=2<3, which does not contradict the one.

Hence exists and m=2

After moving the item, the coefficient of x can be classified and discussed.

exists, the original arrangement has (m-3)x>2+m, let m-3 is not equal to x(2+m) (3-m), so that (2+m) (3-m)=-4, and m is equal to 14/3, so when m>=5 the integer meets the requirements.

Solution: Because of the integer m, the solution set of the inequality mx-m 3x+2 is x -4, resulting in -5m>-10 and m<2Old.

m is an integer less than 2.

Originalization.

m-3)x>m+2

If the solution set is x<-4, we know that m-3<0 is m<3, then x<(m+2) (m-3).

m+x)/(m-3)=4

The solution is that m=14 3 contradicts m<3.

So there are no m-values.

The speed set in still water is xkm h

then there is 66 (x+3)=48 (x-3).

Solution x=19

So the speed of the ship in still water is 19 km h

66/(x+3)=48/(x-3)

Simplified to 18x=342

So the speed of the ship was 19 h

According to the same time, the equation is 66 (3+v)=48 (v-3), and the velocity of the ship in still water is 19 if v=19 is obtained

Solution: Let the hydrostatic velocity be xkm h

66 (x+3)=48 (x-3) according to the title

Solution x=19

Solution: (1) Assuming that the purchase price is m yuan, there is according to the topic.

75 x = 40 A: The purchase price is 40 yuan.

2) Profit w=y(75

w=(20+4x)(20-x)

w=-4x^2+60x+400

w=-4 (It can be known that the image of the god orange w and x is a parabolic line.)

When x=, the function has a maximum value of w=625

Answer: **The maximum profit obtained per day during the period is 625 yuan, and the price should be reduced.

1 solution: set the purchase price to z yuan.

From the meaning of the question: solution: z=40

A: The purchase price is 40 yuan.

2 solution: w=y(75

w=(20+4x)(20-x)

w=-4x^2+60x+400

w=-4 (w is less than 625 yuan for the sock beam.)

That is, when x is equal to, w has a maximum value of the band, w=625

A: The maximum value is 625 yuan.

The ratio of side lengths is 4:5, and the ratio of perimeters is also 4:5, so the circumference of two squares is 36 (4+5) 4=16, 36 (4+5) 5=20, and the side lengths are 16 4=4, 20 4=5, and the sum of the areas is 4 4+5 5=41

x-5=0, x=5

x-4=0, x=4

The zero point values are 4 and 5

When x<4, the original formula =-(x-5)+[x-4)]=-x+5-x+4=-2x+9

When 4 x < 5, the original formula =-(x-5)+x-4=-x+5+x-4=1 When x 5, the original formula = x-5+x-4=2x-9 algebraic formula |x-5|+|x-4|The minimum value is: 1

After reading the title, the group owner worked very hard and typed out the text of the question stem of the question, but it did not match the display of, **It looks like it should be the original picture, there is no problem, the point p is in the second quadrant, and the coordinates of the point p cannot be (2,2),

(x-2)/2006+x/2007+(x+2)/2008=6

Because 6 = 2 + 2 + 2 = 2 * 2006 2006 + 2 * 2007 2007 + 2 * 2008 2008

So (x-2) 2006+x 2007+(x+2) 2008=2*2006 2006+2*2007 2007+2*2008 2008

The right side is moved to the left side.

x-2-2*2006)/2006+(x-2*2007)/2007+(x+2-2*2008)/2008=0

Tidy up in parentheses.

x-2*2007)/2006+(x-2*2007)/2007+(x-2*2007)/2008=0

That is, (x-2*2007)*(1 2006+1 2007+1 2008)=0

So x=2*2007=4014

1. The quadratic function passes through the point p(2,2), that is, y=2 when x=2, so a=1 22, a(t,1 2t 2), ad is parallel to the x-axis, and the ordinate of d is the same as a, d(-t,1 2t 2).

ABCD is a parallelogram, the abscissa of b is T+1, and the coordinate of C is (-T+1,1 2(T+1) 2).

3. ABCD is a parallelogram, and the area of the triangle EAB is 1 4 of the parallelogram, and the area is 1 4*2T*(1 2(T+1) 2-1 2T 2)=1 2T 2+1 4

1. Substitute the p point into y=ax*2 to get a=

2. Substituting the abscissa of a into the equation to get a(t,, because AD is parallel to the x-axis, so the ordinate of d is equal to a, and substituting the equation to get the coordinates of d is (-t,.

3. The length of AD is 2T, and the coordinates of B are (T+1, and the corresponding height of AD is the ordinate of B minus the ordinate of A, which is equal to T+

The area of the parallelogram is calculated as: 2t*(t+ = 2t*2+teab is 1 4 of the parallelogram, which is equal to:

4. Omitted

(1) Bring p(2,2) into y=ax to get 2=2a, so a=1

2) The abscissa of A is t, so the abscissa of D is -t, so AD=2T, so the abscissa of C is 2T-(T+1)=T-1, and the abscissa of B is substituted into Y=X to get the ordinate of B is T+1, because AD is parallel to the x-axis The ordinate of C is equal to the point B, and the quadrilateral ABCD is a parallelogram, so CB is parallel to the X-axis, so the ordinates of C and B are equal, so C(T-1, T+1).

Kindly remind me that you have to do math problems by yourself, so many small problems, I don't believe you can't do a single one, do more math problems, you can activate your thinking, improve your ability to do problems and the ability to use theorems.

(1) Substituting (2,2) into the analytic formula of the function.

4a=2a=1/2

Many students have already calculated the first two questions. I'll just do the last two questions. The key is to understand my analytical thinking. It will help you.

Mathematics in junior high school is relatively basic, and it is generally possible to improve math scores by doing more practice.

(1) s=-2x²+32x

2) The area is maximum when x=-32 (-4)=8 (m).

The maximum area is (-32x32) (8) = 128 (square meters).

The selling price x the auction price of Jinqing is x-40

5x-5(x-40)=8x*85%-8(x-40)Two hailstones annihilate people's reputation and envy x

2+x+x+x=128