Hurry and beg for math questions 20

1.The denominator is 75, and the simplest fraction means that the denominator and the numerator are not coprimitous, i.e., there is no common divisor except 1. The divisor of 75 is , except for 1 and 75, that is, between 1 75, everything that has a divisor cannot form the simplest fraction with the denominator 75.

The answer upstairs should be right.

2。Answer: 5 11 2007 and 1 2008 Problem solving:

2007 and 1 2008

The third question is not very clear, please elaborate.

and for 20

Or. 2007 2008 * 2009 = (1-1 2008) * 2009 = 2009-2009 2008 = 2007 and 2007 2008

Upstairs there was a slight deviation in the results, hehe.

There are many answers, and I won't list them all.

1.If the value of the denominator 75 is 0 1, the numerator is 1 74 for a total of 74 fractions, and the requirement is the simplest fraction, because the denominator is 75 5 5 3, the numerator can not be a multiple of 3 for a total of 24, nor can it be a multiple of 5 for a total of 14, and the same is a multiple of 3 and 5 for a total of 4, so the number of numerators is 74 24 14 4 40 specifically 1 75 2 75 4 75 75 8 75 11 75 13 75 14 75 16 75 17 75 19 75 22 75 23 75 26 75 28 75 29 75 31 75 32 75 34 75 37 75 38 75 41 75 43 75 44 75 46 75 47 75 49 75 52 75 53 75 56 75 58 75 59 75 61 75 62 75 64 75 67 75 68 75 71 75 73 75 74 75

Summation: Sum of numerators (1 2 ...74）×74/2 -3×（1＋2＋..24）－5×（1＋2＋..14) 15 (1 2 3 4) 75 20 Therefore, the sum of the fractions is 20

1.Generations x = a get (4b-a) a = a -2b

0 b 2+4b>=a to find the minimum value of b 2+4b, then a is less than or equal to -4

A broken d3 heart, hello :

Solution: AOD and ACD are two triangles of the same height with d as the vertex and AO and AC as the base, and S aod:s acd=1:3, ao:ac=1:3 (the ratio of the area of the two triangles of the same height is equal to the ratio of the two bases corresponding to the two heights).

ao:oc=1:2.

ad bc, ado= cbo (two straight lines are parallel, the inner angles are equal).

aod = cob (equal to the vertex angles), aod cob (two angles corresponding to two equal triangles are similar).

s aod:s cob = (ao 2):(oc 2) (the area ratio of similar triangles is equal to the square ratio of the corresponding sides).

s△aod:s△cob=1:4.

This problem is to use similar triangles to evaluate, and the key is to determine which two triangles you want to prove to be similar. For example, in this problem, you want to find the value of s aod:s cob by getting aod cob to get s aod:

s cob=(ao 2):(oc 2) to establish a relationship between known and demanded. To prove that two triangles are similar, we usually have the following 5 methods:

1) Definition method: two triangles with equal corresponding angles and proportional sides are similar;

2) Parallel method: a straight line parallel to one side of the triangle intersects with the other two sides (or extension lines on both sides), and the triangle formed is similar to the original triangle;

3) Decision theorem 1: two triangles corresponding to two equal angles are similar;

4) Decision theorem 2: the two sides correspond proportionally and the angles are equal, and the two triangles are similar;

5) Decision Theorem 3: Three sides correspond to two proportional triangles that are similar.

In this problem, we get the aod cob. by using "two angles correspond to two triangles that are equal to each other".

The key to solving math problems is to be good at summarizing and abstracting from the concept in a timely manner, and be able to draw inferences from one another.

<1< <2 can be obtained: in the function of ax 2+3x+b=0, f(1) is greater than 0 and f(2) is less than 0

We get a less than 0, a+b greater than -3, 4a+b less than -6 for the function ax 2+4x+b: f(-2)=4a+b-8, f(1)=a+b+4

From a+b greater than -3 and 4a+b less than -6, we get: f(-2) is less than 0 and f(1) is greater than 0

So -2

Didn't finish the previous equation? Does missing equal to 0?

y=kx=b???Question it)!

1. First bring the two points A and B into Y to get K, B!

2. Let the coordinates of p be (x,y).

Then write the straight line where AP and PB are located! p(x,kx+b) is an isosceles triangle, and the length of ap and bp is calculated according to the equality of the two sides and between the two points, which is represented by x.

ap=bp=>p's coordinates!

3, the third question, alas, you can know how to do it by drawing by yourself (I'm a little irresponsible, but there are a lot of words to write, even if I write the steps, you don't necessarily understand)!

Set a1, a2 ,..., an is a set of base vectors in n-dimensional linear space v, so any vector in linear space can be linearly expressed by this group of vectors, if you want to find the vector with the most 0 in this space (of course, it cannot be all zero), the coordinate vector of course satisfies the condition, it has only one non-zero component, let's.

ei=[0,0,…0,1,0,…,0]

is a vector with the ith component of 1 and the rest of the vector of 0, due to a1, a2 ,..., an is the base, so there must be constants k1, k2 ,..., kn makes.

k1×a1+ k2×a2+…+kn×an= ei

Let a=[a1 a2 ....an] is ,... by a1, a2, an is the matrix of columns, x=[k1,k2,..., kn] t, then it is obtained from the above equation.

ax= ei

This is a system of nth-order linear equations, which can be obtained by Gaussian elimination or trigonometric decomposition to find k1, k2 ,...,kn.

First of all, to answer your next question: [0 1] and [1 0] are not in this space, because you are giving a 2-dimensional vector that cannot form a basis, only a vector of the form [2k, 3k] (k is an arbitrary constant) in this space (collinear), and neither [0 1] nor [1 0] can be expressed as a multiple of v = [2 3].

To answer your starting question: if you don't give r n space, this is better, set a1, a2 ,..., an is an abstract element, because it is a set of base vectors, so all elements of dimensional linear space can be linearly expressed by this set of elements, of course, a1, a2 ,..., an is also in this space, and this group of elements corresponds to the coordinate vector of r n space, i.e.

a1=1×a1+ 0×a2+…+0×an

a2=0×a1+ 1×a2+…+0×an

an=0×a1+ 0×a2+…+1×an

At this point, any of these vectors will meet your requirements.

Suppose you already have the following algorithm:

1.iszero(a) to determine whether subspace a is 0.

2.intersect(a, b) to find the intersection of two linear subspaces a, b.

3. span(v1,..vk), can find vectors v1, v2 ,..A new space for VK.

4.select(a) can take out an arbitrary vector from space a.

Let s = be all coordinate vectors, e.g. e1=(1,0,..0)。

Let a be the subspace you already know.

Then your algorithm is.

for k = 1 to n do )}

The above algorithm starts from the coordinate axis (because the vector on the coordinate axis has n-1 0 yuan, and there is only one non-0) and finds.

It's intersecting with your subspace. If the intersection is not 0, then any vector on this axis is the result of your request.

Then consider the coordinate subplane of the two axes (there are two non-0 elements in this case) and continue the process.

The fourth floor is right.

No matter what kind of linear space it is, as long as it is finite, the combination of vectors in it is inside this space (it seems that you only say that the dimension is high, but you also don't say whether it is finite...).

On the premise of finite dimensionality, 1, if you know a set of bases for this space, then the dimension of space is known first;

2. How do you define the vector with the most "0" you need? Is it the matrix coefficient of the basis vector in Euclidean space? If so, then you take the known set of bases by transforming the matrix into something like (1,0,0...)

The base is perfectly fine.

3, the back is the same as the 4th floor.

This is the most basic knowledge of linear algebra in mathematical form.

Remember that the linear space generated by the row vector is w

Remember: The linear space generated by the b row vector corresponds to the algebraic complement space v1, then s1 = w v1 is still a linear space. The elements in this space are related to the column vector in a, but not to b.

Note that the linear space generated by the c row vector corresponds to the algebraic complement space of v2s2 = s1 v2 and the elements in this space are related to the row vector in a, but not to b and c.

Remember that the linear space generated by the d row vector corresponds to the algebraic complement space of v3s3 = s1 v3 and the elements in this space satisfy the correlation with the row vector in a, and the correlation with b, c, d.

If s3 calculates yes, there is no solution.

It is recommended to use MATLAB to do this, if the amount of computation is relatively large.