Ask questions about basic inequalities in the first year of high school, and questions about basic i

Year 1: A Year 2: (1+a)*a

Year 3: (1+b)*(1+a)*a

It can be obtained: (1+x)*(1+x)=(1+a)*(1+b) i.e.: x squared + 2x + 1 = a + b + ab + 1 ( formula ) 1) when a = b, there is x = (a + b) 2;

2) When a is not equal to b, substitute x=(a+b) 2 into obtain: 1+2(a+b)+(a2) squared + (b 2) squared》 1+2(a+b)+ab 2; ( formula).

and because of 0ab 2;

So the formula "1 + (a + b) + ab 2 + ab 2 = 1 + (a + b) + ab so x<(a + b) 2

To sum up: x<(a+b) 2 or x=(a+b) 2 remember to put the two answers together, I won't hit the less than or equal sign).

The second year's output = a*(1+a).

The output of the third year = [a*(1+a)]*1+b).

Then there is 1+x)*(1+x)=(1+a)*(1+b), that is, 1+x, 2+2x=1+a+b+ab

1), when a=b, there is x=(a+b) 2;

2) When a is not equal to b, substitute x=(a+b) 2 into 1+x 2+2x=1+(a 2) 4+(b 2) 4+a+b+a+b=1+2(a+b)+(a 2) 2+(b 2) 2>1+2(a+b)+ab 2; (z)

and because of 0ab 2;

So Z 1+(a+b)+ab 2+ab 2=1+(a+b)+ab, so x<(a+b) 2

In summary: x<=(a+b) 2 .

Basic inequalities are an important part of high school mathematics, which is of great significance for the cultivation of students' mathematical thinking ability and problem-solving ability. In the following 700 words, I will introduce the topic of basic inequalities in the first year of high school.

Question: Verification: For any positive real number a,b, there is (a+b) 2 >=ab).

Answer: According to the definitions of arithmetic mean and geometric mean, we can get (a+b) 2 >=ab). This is because the arithmetic mean is always greater than or equal to the geometric mean.

So, for any positive real number a,b, the inequality (a+b) 2 >=ab) holds.

Problem: It is known that a, b, and c are positive real numbers, and abc = 1 is satisfied. Verification: a + b + c >=3.

Answer: We can use the mean inequality to solve the problem. According to the mean inequality, we know that for any set of positive real numbers, the arithmetic mean is greater than or equal to the geometric mean.

So, we have: (a + b + c) 3 >=abc). Since abc = 1, (abc) = 1.

Thus, we get (a + b + c) 3 > = 1, i.e., a + b + c > = 3. So, for any positive real number a, b, c, inequality a + b + c >=3 satisfying abc = 1 holds.

Problem: It is known that a, b, and c are positive real numbers, and abc = 1 is satisfied. Verification: AB + BC + CA <=1 3.

Answer: We can use the Cauchy-Schwarz inequality to solve the problem. According to the Cauchy-Schwarz inequality, we know that for any set of positive real numbers, there is (a 2 + b 2 + c 2)(1 2 + 1 2 + 1 2) >a + b + c) 2.

Substituting the conditions in the problem into the inequality, we have (a 2 + b 2 + c 2)(1 + 1 + 1) >a + b + c) 2. Simplify to give 3(a2 + b2 + c2) >a + b + c) 2. Simplify again to get 3(a2 + b2 + c2) >1.

Since a, b, and c are positive real numbers, the auspicious modulus is a 2 + b 2 + c 2 > = ab + bc + ca. So, we have 3(ab + bc + ca) >1, i.e., ab + bc + ca < = 1 3. So, for any positive real number a, b, c, inequality ab + bc + ca <=1 3 that satisfies a + b + c = 1 3.

The above are the questions and answers about the basic inequality of the first year of high school. Through the practice of these problems, students can master the application and problem-solving skills of basic inequalities, and improve their mathematical thinking ability and problem-solving ability. At the same time, it can also cultivate students' logical thinking and mathematical proof ability.

Hope the above is helpful to you.

Solution: Because the state solution set of ax 2+bx+c<0 is {sail spike x xn}(m0 and a<0 to get c<0

Solve the equation cx 2-bx+a=0

x1+x2=b/c x1x2=a/c

The solution is x1=-1 m x2=-1 n or x1=-1 n x2=-1 m

So cx 2-bx+a>0 can be written.

c)x^2+bx-a<0

The solution set is (-1 m, -1 n).

cx²-bx+a>0

Divide x on both sides of the equation at the same time

c-b/x+a/x²>0

a(-1/x)²+b(-1/x)+c>0

Because the solution of ax +bx+c<0 is xn

So the solution of ax +bx+c>0 is m0, which must be satisfied.

m<-1/x-n>0

So the solution set of -1 m0 is.

1.Equivalent to (3x 2)[(x-1)(x 2+x+1)]<=0, equivalent to (3x 2)(x-1)<=0 (because the discriminant that is reduced to 0 is always greater than 0), x<=1 is greater than or equal to -2 3

2.The mean inequality x 2 + 1 (4x) + 1 (4x) is greater than or equal to 3 2 * (2 to the 3rd power).

3.As with the above, the form of the complement mean inequality is +2 and then -2, and the above equation is greater than or equal to 2*, and the root number is [(2x-2)* 1 (x-1)]+2, and the minimum value is 2 times, and the root number 2+2 is established, and when the equal sign is established, 2x-2=1 x-1 can be solved.

4.If the second inequality is greater than 1 and less than 4, then the first inequality is x greater than or equal to 1 and less than or equal to a. Thus, a is greater than or equal to 4

How did you learn math in junior high school? I know how to do it in junior 3, hey, think about it yourself.

solutions, discussed separately.

When, a>1, b is definitely greater than 1(b>a). So just remove the absolute value and get the a=b contradiction.

When, b<1, a must be less than 1(b>a). Therefore, the absolute value is removed and solved with a negative sign, and the a=b contradiction.

So it can only be a<1, b>1Remove the absolute value, add a negative sign on the left, and remove it directly on the right, and solve, 1 a+1 b=2

b=a/（2a-1）

So 2a+b=2a+a (2a-1).

Then find the minimum value of f(a)=2a+a (2a-1), and you know 0 and leave the rest to yourself.

To do the FX image, combine the images and conditions to push out 0 A1 and B>1, and at the same time get 1 A+1 B=2

Because a and b are positive real numbers, b a)x +a b)y 2*root number[(b a)x *a b)y ]2 |xy| ≥2 xy；

the same way(c a)x +a c)z 2(xz);

c/b)y² +b/c)z² ≥2（yz）;

Add the three formulas to get.

b+c）/a]x² +a+c）/b]y² +a+b）/c]z² ≥2（xy+yz+xz）

The three test takers on the left are mean inequalities, and the two cars are paired! Just add it up! This question is mainly a comprehensive method!

Answer: 5 6

You can count the process yourself.

At -2 m 2, f(x) 0 is constant, we can think of f(x) as a primary function about m, and write it in another form as f(x)=mx+mx 2+m-6 0

In the first type, when m=0, -6 0, constant holds.

The second type m (0,2) is a primary function with a slope greater than 0, as long as the maximum value is less than 0.

There is a maximum value at m=2, so -2 x 1

In summary, it is (0,1).

The third type m [-2,0) is a primary function with a slope less than 0, and the maximum value is less than 0.

m=-2 has a maximum. This is Heng established.

So to sum up.

It can also be regarded as a function after a function, directly bring m=plus or minus 2 into less than 0 to stand, and then take the intersection.

Transform the principal method.

Remember g(m) = m(x 2 + x + 1)-6, when -2 m 2, g(m) < 0 so g(2) < 0 and g(-2) <0

That is, 2(x 2+x+1)-6<0 and -2(x 2+x+1)-6<0 are solved to -2, so the range of x is (-2,1).