Math masters, please solve the problem 1 4 methods .

Method 1 If the speed of the car is 3/2 of the speed of the motorcycle, the distance of the car at the time of the encounter is 3/2 of the speed of the motorcycle

The car traveled on a rectangular road.

3 (2 3) 3/5

This rectangular road is complete.

4 (3/5 1/2) 40 km.

Method 2 When we met, cars traveled more than motorcycles.

4 2 8 km.

Motorcycles are gone.

8 (3 2) 2 16 km.

The car is gone.

8 (3 2) 3 24 km.

The rectangular road is the full length.

16 24 40 km.

Method 3: If the motorcycle travels x kilometers and the car travels 3/2 x kilometers when they meet.

3/2 x 4 x 4

3/2 x 4 4

1/2 x 8

x 16 cars traveled 3/2 x 3/2 16 24 km.

The rectangular road has a total length of 16 24 40 km.

Method 4: Set up a rectangular road with a total length of x kilometers.

3 (2 3) x 1/2 x 4

3/5 x 1/2 x 4

3/5 x 1/2 x 4

1 out of 10 x 4

x＝40

From the meaning of the title, it can be seen that the speed of the motorcycle is greater than the speed of the car, and if the two meet at point B, that is, the car directly reaches point B, and the motorcycle passes through point C and then to point B.

Solution 1: Let the total distance be s and the speed of the motorcycle be v, then the speed of the car is 2 3v series of equations: s 2 + 4 = vt

s/2-4=2/3vt

t is the same, then s=40 can be solved

Solution 2: Let the motorcycle speed be V

vt-2/3vt=8

If vt=24 can be solved, then s=<24-4>*2=40 Solution 3: The speed of the car can be set.

The sum of the two sides of the triangle is greater than the third side.

x+2x-1>5x-3

x<1

Because it is an isosceles triangle, there are two sides that are equal.

Assuming x=2x-1,=>x=1>2 3, it does not meet the requirements.

Assuming 2x-1=5x-3,=>x=2 3, then 2x-1=5x-3=1 3, the sum of the two sides is equal to the third side, does not match.

Therefore: x=5x-3 =>x=3 4, at this time 2x-1=1 2 so the perimeter is 3 4+3 4+1 2=2

Let's ...... in three cases

Just be clear that the triangle must exist and must satisfy that the sum of the two sides is greater than the third side.

As for the isosceles triangle, it only needs to be satisfied that the sum of the two waists is greater than the bottom edge.

x=1, so the three sides are 1, 1, and 2 respectively

But 1+1=2

So there is no triangle, ...... is discarded

x=3 4, so the length of the three sides is 3 4, 1 2, 3 4, so it fits the topic, and the circumference is 3 4 + 3 4 + 1 2=2x=2 3, so the length of the three sides is 2 3, 1 3, 1 3, but 1 3 + 1 3 = 2 3

So the triangle doesn't exist.

In summary, only the second type holds, and the circumference is 2

1。。x=2x-1

x=15x-3=2=1+1, not compliant.

2...2x-1=5x-3

x=2 32 3,1 3,1 3,non-compliant.

3...5x-3=x

x=3 4 yes.

then the three sides are 3 4 3 4 , 1 2

The circumference is 2

Isosceles triangle with 2 equal sides.

Let x=2x-1, then x=1, and the three sides are 1,1,2, which is not satisfactory.

Let x=5x-3, then x=3 4, and the three sides are 3 4, 1 2, 3 4, which meets the requirements.

Let 2x-1=5x-3, then x=2 3, and the three sides are 2 3, 1 3, 1 3, which is not satisfactory.

So the three sides are 3 4, 1 2, 3 4, and the circumference: 2

1）g(x)'=1/2√x

g(4)'=1/4

So the slope of the tangent of g(x) at (4,2) is 1 42) and the tangent is .

y=1/4x+b

Bring (4,2) in, 2=1+b, so b=1

So the tangent equation is y=1 4x+1

Bring (4,2) into f(x), k = 1 4

4) Lianli (4, 2).

The circumference is 2 little brother (sister) This has to be calculated by yourself, the two waists of the isosceles triangle are equal, and now I don't know which two are the waist, then try, verified (the result of the calculation is substituted into the sum of the two sides is greater than the third side), calculate it again! Step up your game!

1）g(x)'=1/2√x g(4)'=1 4 So the tangent slope of g(x) at (4,2) is 1 4

2) Let the tangent be y=1 4x+b bring (4,2) in, 2=1+b, so b=1 so the tangent equation is y=1 4x+1

3) Bring (4,2) into f(x), k=1 44) Polaroid (4,2).

1) The slope is x=4,g'Valid values.

2) The slope of the point (4,2) is obtained from (1).

Straight line equations: ...

3) Substituting the point (4,2) into f(x) to find k

4) System of equations.

g(x) = root number x

f(x)=kx^2

Finding the solution to this system of equations is the intersection point.

1): is to find the reciprocal of g(x), and then substitute x=4 into its reciprocal, and the slope is equal to 1 4 (2): the point slope oblique:

y=1 4x-1 (3) substituting the point directly into f(x) to obtain, k=1 8 (4) two-formula sub-series equations, solve: (4,0),(2,-3 2).

Set the bucket height to x

8*15 = x*5 and x=24

Solution: It is known that there are three peaches, and it can be considered that each person has 1 3.

Now assuming that everyone gets apples, then a total of 3 100 = 300 is needed, which is more than the actual 100 (300-100 = 200), because the people who divide the peaches are regarded as the people who divide the apples, and each person has more (3-1 3 = 8 3), so there are 200 (8 3) = 75 people who share peaches, 75 3 = 25 peaches, and 100-25 = 75 apples.

Column: (3 100-100) (3-1 3) 3=25, 100-25=75

Equation solution. If there are x people who divide peaches, then there are (100-x) people who divide apples, 1 3*x+3(100-x)=100, x=75

There are peaches 75 3 = 25 and apples 100-25 = 75

Congratulations upstairs the answers are all wrong, should be the other way around, 25 peaches, 75 apples. Embarrassment.

If there are x peaches, then 100-x apples.

3x+（100-x）/3=100

x=25100-x=75

So.

x peaches, 100-x apples.

x*3+(100-x)/3 = 100

solve, x=25

So 25 peaches and 75 apples.

Set a total of x peaches and 100-x apples.

x*3+100-x/3=100

9x+100-x=300

8x=200

x=25, so 100-x=75

So there are 25 peaches and 75 apples

If there are x peaches, then 100-x apples.

3x+(100-x)/3=100

8x/3=200/3

x=25100-x=75

25 peaches and 75 apples.

There are three apples per person, and 100 can be just divided?

The straight line is two sections of infinite extension, and the straight line can be drawn through the AB point; The ray CD is based on the point C, and the ray is drawn through D;

Ray AD is based on A, and the ray is drawn through point D, and it goes without saying that it is connected to BC 4Just follow the prompts and you're good to go.

I don't know if you understand that?

The straight line AB is to connect the two sides of A and B, the ray CD is not out of C, the ray AD is the beginning of A is not out of D, and the connection of BC is not out of both sides, draw the A end of AB a longer, and the D end of CD is also a little longer, and it will intersect and mark E on the point

Solution: x 2 + y 2 = 1, the center of the circle is (0, 0) and the radius is 1x 2 + y 2-8x + 12 = 0

It can be seen that (x-4) 2+y 2=4, the center of the circle is (4,0), the radius is 2, and the center of the moving circle is ( x,y).

Then the trajectory equation for the center of the moving circle is: (x 2+y 2) (1 2)-1=[(x-4) 2+y 2] (1 2)-2