Ladies and gentlemen, big math brothers. How to do this question

Fang actually uses the representation.

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Proof: Because 2 a = 5 b = 10 c≠1, a, b, c ≠ 0.

Multiply both sides of 2 a=10 c by 1 a to the power of .

2^a)^(1/a)=(10^c)^(1/a)2=10^(c/a)

Multiply both sides of 5 b = 10 c to the power of 1 b.

5^b)^(1/b)=(10^c)^(1/b)5=10^(c/b)

Multiply the two formulas to get it.

2×5=10^(c/a)×10^(c/b)10=10^(c/a+c/b)

The base number is the same, so the number of times is equal, and it can be obtained.

1=c/a+c/b

Multiply both sides by ab to get.

ab=ac+bc

If my is helpful to you, please choose it as a satisfactory answer in time, thank you

is used to indicate how many powers, then the original question is deformed as:

Known: 2 a=5 b=10 c, verified: ab=ac+bc.

Proof: Because 2 a = 5 b = 10 c≠1, a, b, c ≠ 0.

Multiply both sides of 2 a=10 c by 1 a to the power of .

2^a)^(1/a)=(10^c)^(1/a)2=10^(c/a)

Multiply both sides of 5 b = 10 c to the power of 1 b.

5^b)^(1/b)=(10^c)^(1/b)5=10^(c/b)

Multiply the two formulas to get it.

2×5=10^(c/a)×10^(c/b)10=10^(c/a+c/b)

The base number is the same, so the number of times is equal, and it can be obtained.

1=c/a+c/b

Multiply both sides by ab to get.

ab=ac+bc

Knowing 2 a=5 b=10 c, verify: ab=ac+bc=c(a+b) prove: let 2 a=5 b=10 c=k

Then 10 (ab) = 2 (ab) *5 (ab) = (2 a) b * 5 b) a = k b * k a = k (a+b).

10 c(a+b) = (10 c) (a+b) = k (a+b) so 10 (ab)== 10 c(a+b) so ab = c(a+b).

Let the known be equal to w, then use the logarithmic property to get a=log2(w), b=log5(w), c=log10(w), and then use loga(b)=1 [logb(a)], to get it.

1 a=logw(2), 1 b=logw(5), 1 c=logw(10), so we get 1 a+1 b=1 c, and then multiply both sides by abc, remember to prove ab=ac+bc

Let 2*a=5*b=10*c=k, then a, b, and c are all represented by k, and then substitute ab=ac bc to verify.

According to the full probability formula: whichever pocket is drawn, the probability is 1 3

One way is to use the combination number (there is a solution, no repetition, but the result is incorrectly calculated), and the other way is to directly use the probability to find:

1. The probability of taking out each bag is 1 3,2, the probability of taking out two white balls in each bag = the probability of taking out the white ball for the first time The probability of taking out the white ball for the second time, the probability of taking out the white ball in bag A is: 3 4 2 3=1 2, the probability of taking out the white ball in bag B is: 3 5 2 4=3 10, and the probability of taking out the white ball in bag C is:

2 4 1 3 = 1 6,3, the probability of finally taking out two balls is:

What should I do if the dormitory in front of me takes over the street problem? Well, this has to be seen, if it is a question when you are a child, it is easier to do, if the question in junior high school is more difficult to do, it has more steps, and when you get to high school, it is even more topical, I don't know too much, it is easy to buy and sell, and it is easiest to remember wrong, so you have to slowly enter his passport step by step, in order to solve his answer.

In fact, math problems are also very easy to make, you just need to follow the steps to do it, it is relatively simple.

Probability problems are sometimes very complex, sometimes very simple, different methods and even different results, there is an absolutely correct method, that is, to find the ratio of the number of combinations to the overall number of combinations.

The first step is to sum up the white and black balls to get 3 + 3 + 2 = 8 white balls and 1 + 2 + 2 = 5 black balls.

The second step is to find the number of all whites, c(8,2) = 28, c(3,2) =3, c(3,2) =3, c(2,2) =1

So all white = 28 - 3 - 3 - 1 = 21

Find all black, c(5,2) =10, all black with a c(1, 2) =0, all black with b c(2,2) =1, and c(2,2) =1

So all black = 10 - 1 - 1 = 8

Find black and white, c(8,1) *c(5,1) = 40, same as a=3 * 1 = 3, same as b = 3 * 2 = 6, same as c = 2 * 2 = 4

So black and white = 40 - 3 - 6 - 4 = 27

Total combinations = 21 + 8+ 27 = 56, of which 21 are all white, so the probability is 21 56 = 3 8

In the second method, the probability of taking a white ball in a = 3 (3+1) = 3 4

The probability of taking a white ball in b = 3 (3+2) = 3 5

Probability of taking a white ball in c = 2 (2+2) = 1 2

The method is (ab + bc + ac) c(3,2) = 3 4 * 3 5 + 3 5 * 1 2 + 3 4 * 1 2) 3

So the probability is 3 8

The results of the two methods are the same, and the answer is definitely 3 8. In particular, the first one is very specific. The second approach is more abstract.

Probability of taking a white ball in a = 3 (3+1) =3 4b probability of taking a white ball = 3 (3+2) =3 5c probability of taking a white ball = 2 (2+2) =1 2 is taken as (ab + bc + ac) c(3,2) =3 4 * 3 5 + 3 5 * 1 2 + 3 4 * 1 2) 3

So the probability is 3 8

1) If you get the nail bag, the probability is 1 3. Then take 2 balls from the armor bag, a total of c(4,2)=6 combinations, of which 2 balls are white with c(3,2)=3 combinations, so the probability of getting 2 white balls is 1 2In this case, the total probability is 1 3 * 1 2 = 1 6

2) If you get bag B, the probability is 1 3Then take out 2 balls from bag B, a total of c(5,2)=10 combinations, of which 2 balls are white with c(3,2)=3 combinations, so the probability of getting 2 white balls is 3 10In this case, the total probability is 1 3 * 3 10 = 1 10

3) If you get bag C, the probability is also 1 3Then take out 2 balls from bag C, a total of c(4,2)=6 combinations, of which 2 balls are white with c(2,2)=1 combination, so the probability of getting 2 white balls is 1 6In this case, the total probability is 1 3 * 1 6 = 1 18

In summary, the total probability is 1 6 + 1 10 + 1 18 = 29 90. I don't know if it's right.

This is a step-by-step question. The probability of each of the three bags being selected is 1 3

After the first bag, the same ball is 2 out of 3 white balls; The second bag is 2 of the 3 white balls or 2 black balls; The third bag is 2 white balls or 2 black balls.

Total: 1 3 *(c c)1 3*[(c +1) c ]1 3*[(1+1) c].

Because love is a single number, there are three cases, i.e., love rolls refers to 1 or 2 or 3. Rent potatoes.

1. When love is 1, it becomes 11111 1=11111, which seems unreasonable.

2. When love is 2, 2 is happy, and it is not happy to accumulate a single place.

3. When love is 3, 3 music, if you accumulate a single place for music, music can only be 5, but the highest 3 3 is greater than 5, which is not suitable.

So, only in the first case can the equation be solved.

First of all, let's look at the last word: music multiplied by love equals happiness, indicating that love is equal to 1 (any number of Wuhui multiplied by 1 is equal to itself).

Let's look at the middle three words: the three words in the middle cover and the answer are multiplied by love (1) are all equal to music, indicating that the middle three words are equal to music.

Finally, look at the first word: love multiplied by love equals happiness, courtship equals 1, so happiness equals 1.

So the original formula 11111x1=11111

ac=bc=2 3, bcd=60°+30°=90°,bc dc, bcd area:

4 2 3 2 = 4 3 (hectares).

2) It is more complicated, and the solution is detailed in the figure below

First of all, the edge of the BCD cd=4 is fixed, so the height of the CD edge of the BCD is the largest, and the area is the largest. That is, the distance from B to the straight line CD is the largest.

Based on this, we find a way to figure out what the trajectory of point b is.

If c and d are fixed, then the trajectory of a is a circle with d as the center and a radius of 2 (the red circle in the figure).

Let's get a few bs at a particular position, and then guess the shape of b's trajectory.

1>A is located at point A1 on Cd, at this time, AC has a minimum length A1C, point B is another vertex of an equilateral triangle with A1c as the edge, vertex B1 is located above Cd, 2>A is located at point A3 on the extension line of Cd, at this time AC has a maximum length, 2+4=6, point B is another vertex of an equilateral triangle with A3c as the side, and the vertex B3 is located above Cd

3. > the general position a, ac and circle d have two intersection points a and a', corresponding to b has two positions, b, b', two equilateral triangles ABC with A'b'c is the same vertex c, and b is farther away'Corresponding b'The CD area is larger.

Obviously, the 4 b points obtained by the stop plane cannot be located on a straight line, and guess that b should be a circle in the trajectory.

4> Consider a special case, A is in the position of A2 in the diagram, A2C is tangent to the circle D, and Da2 and A2C are perpendicular to each other, which is the case of (1) sub-problem, where B2C is perpendicular to C, and B2 is the rightmost point of B. A2DC=60°, Da2 is parallel to A3B3.

5> extend the intersection of DA2 and CB3 at O, according to the parallel line interception of the proportional line segment, ob3=2, OC=4, connect ob, observe two triangles obc, and DAC, ACD= BCD-60°, BCO= BCD-60°, ACD= BCO;

oc=dc,bc=ac(equilateral), acd bco,ob=oa=2.

Therefore, the trajectory of b is a circle with o as the center of the circle and a radius of 2.

6> the highest point of the circle O, B0, corresponds to the largest area BCD

The perpendicular line OA1 of the cd is passed, and B0A1=2+2 3

BCD maximum area = 4 (2 + 2 3) 2 = 4 (1 + 3) (h).

The growth of the factory, the ninth middle school of the spring market, the material of alcohol allergies, forgive you, lalala, the wings of Demacia, the wings of the winged tiger, the brothers, the family, good, good.

Tianyu asked the god of mathematics, how to do this problem?

The answer is 1, where the cd position table in your graph is wrong, af:cf=be:ec=1:2ef parallel ab, triangle efc:triangle abc = (cf:ac) squared = 4:9

Area of triangular FEC = 2

Area triangle aef: triangle fec=af:cf=1:2 answer 1

From the view, it can be seen that the bottom surface of the triangle in the top view is 6 long, the ground height is 4, and according to the Pythagorean theorem, the waist length is 5, so the side area is equal to.

According to the relationship between the equal radius of the same circle and the half-chord length, chord centroid distance and radius, the relevant equivalent relations are obtained.

To be continued. The second question is a bit troublesome to find the maximum value.

I chose to use the derivative to find the maximum.

For reference, please smile.