High school math problems, let me solve them

The process is more difficult to fight, you do what I told you: this problem uses the cosine theorem and the sine theorem plus a simple simplification, sin(a-b)=sinaconb-sinbcona, the sinb on the right, sinc use the sine theorem to the form of sina, you can make about sina, use the cosine theorem to remove conb, cona, and you can get the formula on the left.

It's not that hard to ask questions like this, just look at the form and take the time to simplify it.

To make it easier to write, let's make sina=a sinb=b sinc=c

So the original formula = (a 2-b 2) c 2 = (a + b) (a -b) c 2

Now let's see: a+b=sina+sinb=2sin[(a+b) 2]cos[(a-b) 2].

a-b=sina-sinb=2cos[(a+b)/2]sin[(a-b)/2]

So (a+b)(a-b)=2sin[(a+b) 2]cos[(a+b) 2]*2sin[(a-b) 2]cos[(a-b) 2].

sin(a+b)sin(a-b)=sincsin(a-b) (because sin(a+b)=sinc).

So the original formula = sin(a-b) sinc

f(x)=x^3+3mx^2+nx+m^26f'(x)=3x^2+6mx+n

f(x) has an extreme value of 0 at x=-1

f'(-1)=3-6m+n=0

f(-1)=-1+3m-n+m^2=0

From the above equation, (m-1) (m-2)=0 is solved to obtain m=1 or m=2, and m=1 is substituted into f'(-1) The solution is n=3, and m=2 is substituted into the solution to solve the trapped land n=9, so m+n=4 or 11

Please dismantle the land and sell the ruler to continue to the next question.

The room fee is increased by n steps, that is, the room fee = 200 + 20n yuan.

The number of rooms = 300-10n rooms.

Total income y=(200+20n)(300-10n).

200（n^2-20n-300)

Find the maximum value of y.

For the quadratic function y=ax 2+bx+c (a is not equal to 0) (this is called the "general formula").

If a>0 then the function has a minimum value, when x=-(b 2a), y takes the minimum value, and the minimum value is y=(4ac-b 2) 4a

If a<0 then the function has a maximum value, when x=-(b 2a), y takes the maximum value, and the maximum value is y=(4ac-b 2) 4a

For the quadratic function y=a(x-h) 2+k (a is not equal to 0) (this is called the "vertex formula") if a>0 then the function has a minimum value, when x=h, y takes the minimum value, and the minimum value is y=k

If a<0 then the function has a maximum value, when x=h, y takes the maximum value, and the minimum value is y=k

So, when n = -(b 2a) = - (-20 2) = 10, y has a maximum.

y=-200(10 2-20 10-300)=80,000 yuan.

Let the increase be an ex multiple of 20 yuan, and x is a positive integer.

then it will be reduced by 10x.

(200+20x)(300-10x)>300*20060000+4000x-20x>60000x(x-200)<0

0 is 1 x 199

220≤20x+200≤4180

So raise it to 220 yuan and start the total income to increase.

Solution: Suppose x increase by 20 yuan, and the total income is y yuan.

y=(300-10x)(200+20x)

200x^2+4000x+60000

200(x^2-20x)+60000

200(x-10)^2+80000

According to the maximum value of the unary quadratic function, it can be obtained: when x=10, the function has a maximum value of 80000, so when the room rate is increased to 400 yuan, the total revenue of the room per day is the highest.

If you set up an increase of x yuan, the number of guest rooms will be 300-10 * x 20 = 300-x 2.

The room rate is 200+x RMB.

The income is w=-x 2 2+200x+60000=-1 2(x-200) 2+80000, when w>200*300=60000, the inequality is solved.

0 is raised to more than 200 yuan, less than 400 yuan can be raised. When it is increased by 200 yuan to 400 yuan, the total income is the highest.

If you think this answer is correct, please give it points.

Let the increase be $ x, and the income will be y

Room rate 200+x number of rooms300-10x 20y=(200+x)(300-10x 20)y-300*200>0 The range of x.

If the room rate is increased by 20x yuan, the number of rooms will be reduced by 10x, and the total income will be y.

y=（300-10x）×（200+20x）=-200(x-10)²+80000

Obviously, when x = 10, Y has a maximum value of 80,000, and the room rate is 400 yuan.

The room fee will be increased by X yuan.

Profit = (300-1 2x) x (200 x).

1/2x^2 200x 60000

Therefore, when x=-b 2a=200, the maximum return is 80,000 yuan.

Draw out 160 middle-aged, 20 young people and 20 Xuzhou deficit god old age.

Let t=2x-1, then x=(t+1) 2, substituting the original formula to obtain, f(t)=4*((t+1) 2) 2-1, let t=3, f(3)=15

f(a)=4*((a+1) 2) 2-1=15, and the solution is a=3

3=2x-1

x=2 so.

f(3)=4*2^2-1=15

Let 2x-1=a

then there is 4x 2-1=15

x^2=16

x = 4 or -4

So a = 7 or -9

1）c=√9 - 5 =2，e=c/a=2/3, (3/2)|pf|=(1/e)|pf|, which is defined by the second ellipse of the ellipse: the distance from point p to the left alignment|ph|=(1/e)|pf| ,pa|+(3/2)|pf|=|pa|+|ph|≥|ah|≥|as|, only when the ap left alignment L is equal sign, that is, |pa|+(3/2)|pf|The minimum value is obtained, at this time the ordinate of point p y=1, the abscissa x=(6 5) 5, the minimum value = 1+a 2 c = 11 2.

2) Set the right focus to f2 because |pf| +pa| +af2|≥|pf| +pf2|, only when p, a, f2 are in a straight line, take the equal sign, pf| +pa| ≥pf| +pf2| -af2|=2a - 1-2)^2 + 1-0)^2=√2 ；So |pf| +pa|The minimum value is 6 - root number 2;

When point p is below the ellipse, i.e., at the position of p1 in the diagram, there is a | in the triangle ap1f2p1a|≤|p1f2| +af2|, P1, F2, A three points in a straight line take the equal sign;

So you can get: |p1f| +p1a|≤|p1f| +p1f2| +af2|；

Therefore|p1f| +p1a|The maximum value is |p1f| +p1f2| +af2|=2a +√2=6 + 2；

That is, |pf| +pa|The maximum value is 6 + 2;

1)5x^2+9y^2=45

x^2/9+y^2/5=1

The eccentricity is 2 to 3

There is a uniform definition of conic curves.

3 2pf is the distance from p to the left alignment.

The minimum of PA+3 2pf is when PA is perpendicular to the left alignment.

pa+3/2pf)min=11/2

p(6√5/5,1)

2) The second question is also a skill question.

Pa+PF=PA+2A-PF2 (i.e., the right focal point of the ellipse) PA+PF=6+PA-PF2

Since PA-PF2 AF2 (the difference between the two sides of the triangle is less than the third side), 6-AF2 PA+PF6+AF2

6-√2≤pa+pf≤6+√2

Let the distance from p to the right alignment be mp

Because e=pf mp, and because e=2 3, so 3 2pf=mp so pa+3 2pf=pa+mp=1 3

It's the same with the left focus.

The second question is to set a point, which is a simple solution to the equation, and both points a and f are fixed-point and easy to find.