Inequality short answer questions, basic inequality questions

。Because |x2-4|Definitely greater than or equal to 0 |x2-4|<1 So|x2-4|It must be a positive decimal or 0, so x2-4 is less than or equal to 1 or x2-4 is equal to 0, and we get that x2 is less than or equal to 5, x is less than or equal to the root number 5, or x is equal to plus or minus 2, and bring in |x-2|It follows that a is greater than or equal to the root number 5-2 or a is equal to 4 or 0 and 4 is included in the root number 5-2.

Finally, it is concluded that a is greater than or equal to the root number 5-2 or a=0

Upstairs upstairs is mainly a value range. I'm so tired. After playing, I found that the first floor appeared.

Let's see for yourself. I don't know how to give it to me, I'll elaborate on it.

Solution: Because |x^2-4|<1 was established.

By |x^2-4|<1 got.

3＜x＜√5 -√5＜x＜-√3

Again|x-2|0

2-a＜x＜2+a

There is a topic to know.

3 2-a 2+a 5 or -5 2-a 2+a - 3a 5-2, a - 3-2

The value range of a>0a is (0, 5-2].

Let m = x and n = y, then the inequality becomes: m+n a (m +n), which is equivalent to the inequality: (m+n) grinding a (m +n), chain limb.

Equivalent to the inequality: (m n+1) a [(m n) +1], let p=m n, then the inequality becomes blind and ignited: (p+1) a (p +1), and the arrangement is obtained: (a -1) p -2p + (a -1) 0

For inequalities to be constant, it is required. Discriminant.

That is: 4-4(a -1) 0, we can get: a 2 because, a>0, the original inequality can be established;

So, the minimum value of a is .

I only write about the process, and I don't write about the result, you know it yourself.

1.The original formula is equal to x 2

2 x + 2 x, then you can use the fundamental inequality method, because the product of these 3 equations is constant.

2.The second one is written as 2x*x*(1-3x)*1 2, and then the same basic inequality is used, because the sum of 2x, x, 1-3x is constant.

3.Under the root number is -(2a+b) 2+5ab=5ab-1, so as long as 5ab-1 is the maximum, t is the maximum limb. b=1-2a, substituting 5ab-1 into the formula, only need to find out the maximum value of 5a-10a 2-1 in the interval of (0, 1 2).

1 2 is because b=1-2a, b>0, so a<1 2 I only wrote the process of the chaotic hunger, if there is anything you don't understand, send me a message.

Set up the boss to enter x pieces late.

Option 1: 10*100 + x - 10)*100*= 85x +150

Scheme 2 price: 100 * * x = 89 x.

Hypothesis: Option 2 is better than option 1.

85x +150 > 89 x

The solution is x <150 4, because the missing branch x is an integer x <=37, and when scheme 1 is better than scheme 2, the solution is x >=38, so when the number of entries is greater than or equal to 38, scheme 1 is used.

If the number is less than or equal to 37, option 2 shall be adopted.

Solution: Add x grams of salt.

40+x)/(1000+x)>=10%

x>=200/3

A: Add at least 200 3 grams of table salt to make it a concentration of not less than 10%.

There are 40 grams of salt in the brine.

Then there is water 1000-40 = 960 grams.

To make it reach a concentration of 10%.

Then the water should not exceed 90%.

Later the total amount of brine should be reached.

960 90% = grams.

Add at least grams of table salt.

This problem seems to be a wrong condition, if there is only a, b, c>0 then the original formula can be reduced to.

In this form, the original form will never be greater than or equal to 6.

For a random variable x with a mean of 3 and a variance of 4, the lower bound of p(-2 < x < 8) is determined using Chebyshev's inequality.

2 .The mean is 3, the variance is 4, and the probability distribution of x is symmetrical with the mean 3, and the upper bound of p(x <= 0) is determined by Chebyshev's inequality.

The problem with this problem is that if a=b=c=1, then the original formula = 2+2+2=6<8

Hello, this (c a) b (a b) c (b c) a=c b+b c+a b+b a+a c+c a, the above equation can be greater than or equal to 2+2+2=6 by using the mean inequality

g(x)=ax2+bx+c-x= ax2+(b-1)x+c can be explained by the image of a quadratic function

The two roots of the equation ax2+(b-1)x+c=0 are x1, and x2 is the intersection of g(x) and x.

At 0 "Sail Split x state closed 0, i.e. f(x)-x>0, x< f(x) has not yet sorted out the idea of the second question hall banquet.