Two math problems extra points in 10 minutes 10

cosa = (square of cos2 a - square of sin2 a) = square of cos2 a * (square of 1-tan2 a) = (1 + cosa) 2 * (1-4) thus we get cosa = -3 5 y=sin2x+2cos 2x=sin2x+cos2x+1 = 2*sin(2x+pi 4) +1 under the root number to get the period of 2pi 2=pi

tan 2( 2) = (1-cos ) (1 + cos ) = 2 2 = 4, and the solution gives cos = -3 5

y=sin2x+2cos 2x to deform it into a deformation to find the period.

y=sin2x+2cos^2x=2sinxcosx+2cos^2x=2cosx(sinx+cosx)=2cosx*2sinxcos0=2sin2x

So, the period t=2 2=

tan( 2)=2, then cos = how much?

tan 2( 2) = (1-cos ) (1 + cos ) = 2 2 = 4, and the solution gives cos = -3 5

y=sin2x+2cos 2x to deform it into a deformation to find the period.

y=sin2x+2cos^2x=2sinxcosx+2cos^2x=2cosx(sinx+cosx)=2cosx*2sinxcos0=2sin2x

So, the period t=2 2=

1.The maximum number is A+2 and the minimum number is A-2. [a+2-(a-2)]²a+2-a+2)²=4²=16

2.(-2a +6ab+3b)

3.(a+b) +2b=1, so a+b=0, and because 2b=1, a=,b=.

I hope it will help you, and I wish you a better level. I hope to give a limb banquet.

27 + 9 = 36 36 3 = 12 12-8 = 427 * 3 = 81 81 9 = 9 9-8 = 18 to the third power to get 2 27 9 = 3 2 + 3 = 5 The second problem is solved with this program. . . There are so many solutions ...

#include

main()

int s,i,j,m,n,k,a[8],p=0,q;

for(i=5;i<=9;i++)

for(j=0;j<=9;j++)

if(j==i)continue;

for(k=0;k<=9;k++)

p=0;for(n=0;n<=4;n++)

if(a[n]==i||a[n]==j||a[n]==k||a[n]==m)

if(p==0)printf("%d",s/2);}

27/(9/3)-8=1

The remaining 2 answers used cube roots, which took 20 minutes for brain cells.

3rd cubic root of 27) +9-8=4

8-9 (3rd cubic root of 27) = 5

The programming can be solved very quickly, and it is not easy to calculate if it is calculated.

The first question doesn't seem to be clear?

The second question is like this:

fourfour

One by one. ei ght

Again: four

I tried a lot of numbers in the group, but I couldn't get it right.

What about e i ght?

The four-digit number is: 8532

The five-digit number is: 17064

2。If these 9 letters represent different numbers, then there is no answer, and the single digit does not match.

Question 1:

5's hasn't been counted yet.

Isn't there only one answer to the second question? If there is no restriction on letters, I'll report one:

1. Known 2 (A + B + C) 57 + 74 + 69 200 Then: A (A + B + C) - (B + C) 100-74 = 26 B 57-26 31

C 74-31-43

Answer: The three numbers A, B, and C are each

2. If the number of grain stored in the two warehouses is equal in the title: 400 2 200

Answer: Warehouses A and B each have tons of grain.

1. Let A be x, B be 57-x, C be 69-x, 57-x+69-x=74126-2x=74

x=26, then A is 26

B is 57-x=31

C is 69-x=43

2. The question is not clear, is the number of grain stored in the two warehouses the same or other conditions.

a+b=57

b+c=74

a+c=69

Solution: a=26 b=

2, then the number of grain stocks in the two warehouses, (how?) Same? ）a+b=400

a-30=b+30

a=230,b=170

1.Let the three numbers of A, B, and C x,y,z, x+y=57; ②：y+z=74;③：

x+z=69; 2（x+y+z）=200;Then : x+y+z=100; z=100-57=43 and so on, we can get the value of y,z.

Question 2: Incomplete questions cannot be solved.

1. Let A be x 57-x) + (69-x) = 74 and solve A as 26 B 31 C 43 2. Let A have x tons (x-30) = (400-x-30) and solve x as 230 A has 230 tons and B has 170 tons.

1. A efficiency: 1 8

B efficiency: guess the cover 80% 8 = 1 10

1 Caution 2 (1 Beam Rent 8 + 1 10) = 1 2 9 40 = 20 9 hours.

2,21+21-2) (1-20%-20%-1 3)=40 4 15=150 tons.

The Tabernacle contains 2 (1 8+80% 8) = 20 9 hours of laughter and fighting.

21 2-2) (1-1 3-2 20%) = 150 tons.

1. Set the engineering quantity as the unit 1

A can be completed by doing it alone for eight hours, then A's speed is 1 8;

B can complete 80% of the work alone in eight hours, and the speed is 1 10;Roll down.

The speed of cooperation between A and B is 9 40, so it takes (1 2) (9 40) = 20 9 hours to complete half of the project.

2. If this batch of goods has a total of x tons, then the first time x 5 will be transported, the second time x 21 tons will be transported, and the third time x 5 + 19 will be transported, and x 3 will remain. Column equation: x 5 + 21 + x 5 + 19 + x 3 = x solution to x=150 tons.

Hours Bi Wei.

2.(21 2-2) (1-1 3-2 20%) = 150 tons of land.

The sum of two numbers greater than or equal to 0 = 0

then they are 0, so a= b=-3

2ab/3=-1

The sum of the two terms is 0, i.e.

The root number (2a, 1) 0 is a=1 2 b+3 squared 0, so b=-3

The final result is 1

According to the characteristics of the square and the characteristics of the root number, it is judged that 2a-1 is a negative number, or a zero. Observations show that there are two sets of solutions:

Solution 1: b is -2 and a is 0, then 2ab is zero, and the fifth root of 2/3ab is also 0

Solution 2: A is one-half, B is -3, then two-thirds of ab is -1, and the fifth power is -1

So there are two sets of results, 0 or -1