High School Mathematics Best Value High School Mathematics, Find the Best Value

I'll help you.

I'm pretty good at math, so you can ask me more questions in the future.

First of all, the maximum value problem in high school mainly needs to make use of the form (x-a)2 because the minimum value of a perfect square is 0

Now we're going to construct the perfect square inside your fraction.

1-6x）/3x^2

1-6x+9x 2) (3x 2) -3 is due to (9x 2) (3x 2)=3

In this way, the denominator of the fraction can be fully squared.

3x)^2-6x+1)/(3x^2) -3(3x-1)^2/(3x^2) -3

In this case, the minimum value of the above converted formula is when the perfect square is 0--- i.e. x=1 3

Substituting x=1 3 gives its minimum value of -3

So the minimum value of (1-6x) 3x 2 is -3 and a (1-6x) 3x 2

So a -3

Totally hand-handed, I hope you are satisfied.

Less than a number is constant, which is equivalent to the minimum value of less than this number (since a is smaller than the minimum value of the following equation, it must be smaller than the other values), so the problem is transformed into finding the minimum value of the equation on the right. The solution process is in **:

Because the square of x is greater than 0, multiply 3x squared by a, and move to get 3ax 2+6x-1 less than or equal to 0 When a>0, there is no x on r, and the left side of the equation is always less than 0, so when a<0, there is no real number root formula satisfied, so 36+4*3a<=0, a is less than or equal to -3

Therefore, a is less than or equal to -3

Let a+b=y, then b=y-a, substitute (b-1) envy and filial piety as a+(a-2) b=1 to get the brother's stove.

y-a-1) a+(a-2) (y-a)=1, remove the denominator to get (y-a)(y-a-1)+a(a-2)=a(y-a), to get y 2-y(2a+1)+a 2+a+a 2-2a=ay-a 2, and get 3a 2-a(3y+1)+y 2-y=0,a,y r, so =(3y+1) 2-12(y 2-y).

3y 2+18y+1 0,3y 2-18y-1 0,9- 84) 3 y (9+ 84) 3, so the maximum value of A+B=Y is (9+ 84) 3, and the minimum value is (9- 84) 3.

High school math is the most valuable, high school math, high school math is determined to deal with the state of the sparrow, because many parents don't quite understand, so the best sail is to find this regular teacher for advice.

1 .Obviously, from the properties of the quadratic letter with the number of finch friends, it can be obtained 3 2 Second, the discriminant method For the problem of the maximum value to be obtained, if the known function formula can be transformed into a problem of whether there is a real root of a quadratic equation by appropriate algebraic deformation, then the discriminant formula can often be used to obtain the function.

The most valuable question has always been one of the important contents of high school mathematics in Tanxi, and it is also a hot issue in the college entrance examination. It is comprehensive and has a wide range of applications in production and life. Therefore, the question of finding the best value is which copy of the internal letter code that we have to grasp in high school.

2cos, base shelter y=-4+2sin

x-1)²+y-1)²

2cos -1) +4+2sin -1) 4(sin +cos Bo ruler )-4(5sin +cos) +2630-4 26sin( +

where tan = 1 5).

sin( +1, the maximum value is .

sin( +1, the minimum value sought is .

Use the properties of functions (e.g., primary functions and quadratic functions) 2. Use the parameter commutation method, which is suitable for composite functions and abstract functions, and reduce complex functions to simple basic functions through the method of commutation, and then use the properties of basic functions to find the missing edge solution. 3. The derivative method judges the monotonicity of the function through the judgment of the monotonicity of the function, and finds the derivative through the volt

High School Mathematics Key Topics] Methods for Finding the Maximum Values of Various Ellipses (Chaos 1) 1. Methods to summarize the maximum value problems in the large conic curve chain The solution to the canopy is generally divided into two types: One is the geometric method

The process is correct, but the commutation needs to be careful to define the domain.

sinx=t, so there should be -1 t 1

Quadratic functions are all open down, so there is a maximum, and then the maximum value is found within this range.

The attention of this problem is the range of t after the commutation sinx=t, because -1 sinx 1, so -1 t 1, and then it is the problem of finding the maximum value of the quadratic function.

According to the idea of the landlord:

1 - 2t² +6t

1 - 2[t² -2 * 3/2) *t + 3/2)² 3/2)²]

1 - 2[(t-3/2)² 9/4)]

1 - 2(t-3/2)² 9/2

2(t-3/2)²

2(3/2 - t)²

2(3/2 - sinx)²

It can be seen that as long as (3 2 - sinx) takes the minimum value, this equation can get the maximum value. Obviously, when sinx = 1, 3 2 - sinx takes the minimum value.

Therefore, the maximum value = - 2 * 5

1 - 2(t² -t)

1 - 2[t² -2 * 1/2) *t + 1/2)² 1/2)²]

1 - 2[(t-1/2)² 1/2)²]

1 - 2(t-1/2)² 2 * 1/2)²

2(t-1/2)²

2(sinx - 1/2)²

Obviously, when (sinx -1 2) is the minimum, the equation gives the maximum. That is, when sinx = 1 2 there is a maximum value : 2 * 0 =

If you find the maximum value of f(n)=[30n(n 2-n+20)] [(n+5) 3(n+4) 3] without limiting n, then neither the maximum nor the minimum value exists.

When n is taken in the left attachment of -4, f(n) can be very large (n -4 -, f(n) + When n is taken in the right attachment of -4, f(n) can be very, very small (n -4 +, f(n) -f(n) is taken in the region around n=-5.

If you limit n>0 but take the value consecutively, draw an image, and find that around n=1, f(n) takes the maximum value, and the maximum value is approximately.

It is easier to limit n to a positive integer, and only need to calculate the value of n, where f(1) is the largest, which is 1 45.

When n 5, consider the reciprocal of f(n) g(n) = (n+5) 3(n+4) 3 [30n(n 2-n+20)].

1 30)*n 3+(14 15)*n 2+(311 30)*n+52+[8000+1400*n 2-20400*n] [30n(n 2-n+20)], the fractional part is always less than 1 4, and the integer part is at least greater than 52, and it is an increasing function, so it can be determined that g(n) is an increasing function, so f(n) is a decreasing function.

In this way, it can be determined that f(n) achieves a maximum value of 1 45 when n = 1.

The numerator is the 3rd order of n, and the denominator is of the 6th order, so the extreme value must be determined by n.

Find the maximum value of the function y=3 root numbers (x-5) + 4 root numbers (6-x) and solve the following results:

According to the function, the definition domain of the function can be found as: 5=y=3*sina+4cosa

5*sin(a+b), where tanb=4 3, it is easy to know that when, sin(a+b)=1, y has a maximum value of 5.

Use Cauchy's inequality and you'll be fine...

a:1+a+b=ab<=(a+b) 2 4 gives a+b>=2+2 root number 2b:a 2+1 [b(a-b)]>=a 2+4 [(a-b)+b] 2=a 2+4 a 2>=4

c: Let the intersection point be (x,y).

Get x 2-2x+2=-x 2+ax+b

It is then obtained perpendicular to the tangent of the intersection.

2x-2)*(2x+a)=-1

But I don't know what you're asking for, try it yourself.

The original rock Zheng = a square + 1 ab + 1 a (a-b) + (a-5c) square jujube hand, the fourth term is 0, does not affect the first three items, the middle will be extracted 1 a, the remaining (1 b + 1 (a-b)) with the mean inequality greater than the Ming suspicion is equal to 4 a square, and then with the first term with the mean inequality, greater than or equal to 4Equal sign formation condition a = root number 2

This can be solved with the mean inequality: