Competition questions, urgent, and detailed process

Isn't this taking off your pants and farting?! That's! Put it in a beaker and call it over?! Also scientific, contrary to the principle of life, shit logic. These 2B competition questions really hurt the thinking of Chinese students.

= 9 and 5/8 (offset) (2 and 1/2003 + 2004/2003) 10 9 and 5/8 (offset).

2/2003/2005 10

6011/20030

I don't know if the answer is the simplest, please understand.

It's not good to get a score on it, so you'll just look at it:

2 times 1/2003 times 9 and 5/8 plus the sum of 2003/2004 times divided by 96 and 1/4.

2 and 1 2003 * 9 and 5 8 + 1 and 1 2003 * 9 and 5 8) 96 and 1 4

2 and 1 2003 + 1 and 1 2003)*9 and 5 8 96 and 1 4=3 and 2 2003*9 and 5 8 96 and 1 4=6011 2003*77 8 385 4=6011 2003*77 8*4 385=6011 20030

It is better to use equations to solve the problem as follows.

Solution: Assuming the planning time is x days, the speed of each dog per day is y kilometers.

According to the title.

5xy=5x+3x（y+1) ①

3x=60 ②

The speed of 5 dogs per day in the planned days = total distance.

The 5 dogs ran together for a day, which was 5x, and the sum of the speed of the remaining three dogs was 3x, and the time was y-1+2

The number of planned days minus one day for the previous five dogs plus two days for being late.

That is, the right side of the equation is also the total distance.

Running 60km more will save a day and saving this day for 3 dogs and speed.

The solution. x=20

y=4 is calculated by substituting the original system of equations, correct!!

The total distance traveled is in Equation 1.

5xy=5×20×4=400（km)

A: The Eskimos covered a total of 400 kilometers.

Set the speed v, normal time t Eskimos travel a total of vt kilometers.

vt=v+3/5 v(t+1)

vt=v+60+3 5 v(t-60 v), the solution yields v=40 t=4 vt=160

It's too long to write it all down......What needs to be added, you can mention it again)

1. Using the trigonometric identity about the cotangent, it can be assigned.

x= 3 ctga, y= 3 ctgb, and z= 3 ctgc

where a, b, and c are the inner angles of an acute triangle. From the sort inequality, you may want to make a, b, and c in the same order as a, b, and c (the left side will be smaller). Wandering Void Order.

s=a/(b+c) +b/(c+a) +c/(a+b)

In this case, using the convexity of the cotangent function at [0,pi 2], the left >=2 3*s*ctg( (s-a(b+c))*a 2s) =2 3*s*tg((a (b+c))*a 2s).

Then by the sort inequality, the variable within tg is not less than pi 6. It is easy to prove by the convexity, s>=3 2, and it is obtained by grinding purely.

2、n=7。The problem can be transformed as follows (the problem itself is not simplified, but the description is more convenient): there is a matrix of n rows and 5 columns, each row consists of 3 1s and 2 zeros, and any two rows are different.

Find the minimum n such that the matrix must have a 5th-order sub (square) matrix whose determinant is not 0.

Easy to find, n>6. For example, there is a matrix with 5 rows and 6 columns in a column that is all 1. For n=7, look at the column where "1" appears the most.

Easy to verify, there are 5 or 6 1s in the column. It is sufficient to discuss these two cases (the 5th-order subarray under consideration should contain all rows with the column 0).

3、k=17/2。It is easy to prove by the sort inequality and the mean inequality.

a^4+b^4+c^4 >=a^3+b^3+c^3 >=3abc

So you might as well set a>c>b. Then, using the local adjustment method, it is verified that when (a,b,c) is replaced by (c,1,2-c), a 4 + b 4 + c 4-3abc] (a-b) (b-c)(c-a).

decreases. Finally, it is transformed into an extreme problem of a univariate function about c. After writing the expression as a rational function of (1-c), we can find k=17 2.

4. [No].

5. It is easy to prove by tossing and dividing (2 a-1, 2 b-1) =2 (a,b) -1. Thereupon.

q|(2^p-1, 2^(q-1)-1)=2^(p,q-1) -1

But (p,q-1)=1 or p, so it can only.

p|(q-1)

And obviously q is an odd number, so q = 2kp + 1.

I'm just a primary school student, so I can't read this question clearly, and I probably won't be able to do it.

(2) x +4 Constant >> 0

Then when x 0: x +4>5x

x² -5x + 4>0

x-1)(x-4)>0

x<1 or x>4

then 0 x<1 or x>4

When x < 0: x +4>-5x

x² +5x + 4>0

x+1)(x+4)>0

x<1 or x>4

then -14 or -1

This kind of question classification discussion: first look at (x*x+1) This must be greater than 0, and divide the two sides directly by (x*x+1).

The formula can be simplified to (2-x)(3+x)>0

Then discuss: (1) when x>=2.

2) -3<=x<2.

3) X<-3.

Do the math yourself, pay attention to multiply and divide the negative, and use the reverse sign.

Another way to teach you is to thread the needle.

In this way, no matter how long the inequality is, you find the rightmost solution equal to 0, and if you can take the equal sign, it will be a solid point, and then interspersed back and forth, and all the solutions equal to 0 will be strung together, and the one above the axis will be greater than 0, the lower one will be less than 0, and the axis will be equal to 0