Although we are only a senior one, but the winter vacation mathematics teacher let us buy a set of high school inequality test paper to do. Is there such a special set of inequality test paper? I would like to ask you which one is better?

Although we are only a senior one, but the winter vacation mathematics teacher let us buy a set of high school inequality test paper to do. Is there such a special set of inequality test paper? I would like to ask you which one is better?

Buy Longmen topic "inequality", all kinds of inequality types, very easy to use
What are the commonly used inequality formulas in high school?
(a+b)/2≥√ab
a^2+b^2≥2ab
(a+b+c)/3≥(abc)^(1/3)
a^3+b^3+c^3≥3abc
(a1+a2+… +an)/n≥(a1a2… an)^(1/n)
2/(1/a+1/b)≤√ab≤(a+b)/2≤√[(a^2+b^2)/2]
Please refer to the following resources for details
Common problems of mathematical inequality in Senior High School
Let the inequality MX ^ 2-2x-m + 1
1. M = 0, x > 1 / 2
When m ≠ 0, the combination of number and shape is considered
Considering that △ = 4m ^ 2-4m + 4 = (2m-1) ^ 2 + 3 > 3 > 0. There must be a solution. Suppose two are x1, x2
We can know the expression of X1 and X2 by using the root formula
If M > 0, the opening of the function is upward
According to the image, it must be between two
Now you say m controls the direction of function opening If M is less than 0, it means that the opening of the function is downward The absolute value of M is less than or equal to two It means that M is between two and minus two With zero as the dividing line Two cases of M are discussed respectively Using discriminant Using set to represent x Give me some points Mobile party
There is no intersection point between △ 0 function and y = 0, m < 0 function opening downward combination image is that the function is under the Y axis and opening downward (with the highest point) x takes any value, and then combined with M = 0, the value of X is greater than 1 / 2
Interesting problems in Mathematics
There are 100 coins. If you change all the 2 cents into the equivalent 5 cents, the total number of coins will be 73, and then change all the 1 cents into the equivalent 5 cents, the total number of coins will be 33. How many coins are there in the three kinds of coins?
Let 1 point x pieces, 2 points y pieces, 5 points Z pieces
(1)x+y+z=100
(2)x+2y/5+z=73
(3)x/5+2y/5+z=33
(2)-(3)：4x/5=40,x=50
(1)-(2)：3y/5=27,y=45
Z=5
If f (x) satisfies the relation f (x) + 2F (1 / x) = 3x, find the analytic expression of F (x)
I want to know what principle can do this
Why can we do this
f(x)+2f(1/x)=3x
f(1/x)+2(x)=3/x
f(x)+f(1/x)=x+1/x
f(x)-f(1/x)=3/x-3x
So f (x) = 2 / X-X
Because the formula given in the question has f (1 / x), so 1 / X is also in the domain of definition
So when we replace x with 1 / x, the function still holds
The acquired upper form
Given that the set a = {ax ^ 2 + 2x + 1 = 0 has at most one proper subset, how to find the value range of real number a?
There may not be one, if there is one, at most
At most, there is only one possibility
There may or may not be if there is at most one
It means that a has only one element,
If a has more than two elements, then a has at least two proper subsets.
The method of finding function analytic expression in high school mathematics
Let f (x) satisfy f (0) = 1, f (a-b) = f (2a-b + 1) (a, B are real numbers)
Find f (x)
F (a-b) = f (a) - B (2a-b + 1) has the wrong number
Let a, B = x get
f(0)=f(x)-x＾2-x
f(x)=x＾2+x+1
What's the difference between 6 / 1 and 6 * 1
6 / 1 refers to the division operation, 6 / 1
6 * 1 refers to the multiplication operation, 6 × 1
6 / 1 is a false fraction, a number
6 * 1 means multiplication of two numbers, which is a kind of operation? It's a second grade question
What is e (AX + b) equal to
E（aX+b）=E（aX）+b=aE（X）+b
EY=E(aX+b）=aE（X）+b
DY=a^2 DX
E（aX+b）=E（aX）+b=aE（X）+b
Well, that's it, I hope
Equal to a * e (x) + B
Expected formula: e (x) = X1 * P1 + x2 * P2 + X3 * P3 +... + xn * PN
The formula of variance: D (x) = (x1-e) square * P1 + (x2-e) square * P2 + (x3-e) square * P4 +
+Square of (xn-e) * PN
E（ax+b）=aE(x)+b
The meaning of a math problem
We have a good harvest of tomatoes. When we take 3 / 8 of all the tomatoes, it's more than 24kg when we fill 3 baskets. When we finish the rest, it's just 6 baskets. How many tomatoes can we get
（24×3-24）÷1/8
Why do you answer that
Because when we take all 3 / 8, it is more than 24kg when we fill 3 baskets, when we take all 9 / 8, it is more than 3 × 24kg when we fill 3 baskets, that is, 9 baskets are more than 3 × 24kg, and we take 3 baskets + 6 baskets + 24kg in total, so 3 × (3 baskets + 24kg) - (3 baskets + 6 baskets + 24kg) = 9 / 8 of the total, so the total is [(3 × 3 + 3 × 24) - (3 + 6 + 24)] / (1 / 8), that is (24 × 3-24) / (1 / 8)