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Let abcxyz be a real number, prove (A & sup2; + B & sup2; + C & sup2;) (X & sup2; + Y & sup2; + Z & sup2;) ≥ (AX + by + CZ) & sup2; Our teacher said that we can first change it to 4 (A & # 178; + B & # 178; + C & # 178;) (X & # 178; + Y & # 178; + Z & # 178;) ≥ (2aX + 2by + 2CZ) &# 178;). I look very much like B & # 178; - 4ac, so I don't know how to solve it. Thank you··········
∵(a²+b²+c²)(x²+y²+z²)=a²x²+b²y²+c²z²+a²y²+a²z²+b²x²+b²z²+c²x²+c²y²(ax+by+cz)...
It is known that abcxyz are all non-zero real numbers, a & sup2; + B & sup2; + C & sup2; = x & sup2; + Y & sup2; + Z & sup2; = ax + by CZ
Verification: X / a = Y / b = Z / C
A & sup2; + B & sup2; + C & sup2; = ax + by CZX & sup2; + Y & sup2; + Z & sup2; = ax + by CZ are added to get a & sup2; + B & sup2; + C & sup2; + X & sup2; + Y & sup2; + Z & sup2; = 2 (AX + by CZ) a & sup2; + B & sup2; + C & sup2; + X & sup2; + Y & sup2; + Z & sup2; - 2ax-2by + 2CZ = 0 (
Let the solution set of inequality | x ^ 2-4x + m | ≤ x + 3 about X be a, and 0 belongs to a, 2 does not belong to a, then the value range of real number m is a
Who can tell me the following two? Your answers are different
It belongs to a
|0 ^ 2-4x0 + m | ≤ 0 + 3, that is | m | ≤ 3, - 3 ≤ m ≤ 3
2 does not belong to a
|2 ^ 2-4x2 + m | 2 + 3, that is | M-4 | 5, M-4 > 5 or m-49 or MX + 3
|4-8+m|>2+3
|m-4|>5
m> 9 or M5, M-4 > 5 or m-49 or MX + 3
|4-8+m|>2+3
|m-4|>5
m> 9 or M
|x^2-4x+m|≤x+3
When x = 0
|m|≤3
-3≤m≤3
Because 2 doesn't belong to a
Then when x = 2
|x^2-4x+m|>x+3
|4-8+m|>2+3
|m-4|>5
m> 9 or M5, M-4 > 5 or m-49 or M5, M-4 > 5 or m-49 or MX + 3
|4-8+m|>2+3
|m-4|>5
m> 9 or M5, M-4 > 5 or m-49 or M
Y = (3x-1) / (x + 1) x ∈ [0,3) for range
y=(3x-1)/(x+1）=[3（x+1）-4]/(x+1)=3-4/(x+1)
Because x ∈ [0,3), so x + 1 ∈ [1,4)
4/(x+1)∈(1,4]
So the range [- 1,2)
It is known that the solution set of inequality ax-5x2-a < 0 about X is m. if 3 ∈ m and 5 ∉ m, then the value range of real number a is______ .
If 3 ∈ m and 5 ∉ m, the solution set of inequality ax-5x2-a ＜ 0 about X is m, so there are & nbsp; 3 · a-59-a ＜ 0 & nbsp; & nbsp; & nbsp; & nbsp; ① 5 · a-525-a ≥ & nbsp; 0 or & nbsp; 25-A & nbsp; = 0 & nbsp; & nbsp; & nbsp; ②, a ＜ 53 & nbsp;, & nbsp; or a ＞ 91 ≤ a ＜ 25
3x / (the square of X + 4) range, master help pull!
Help tell the specific separation constant steps!
Thank you!
Wutangyazi friends what is the first order reciprocal ah, help explain ah!
Method 1
1. If x is not 0
Y = 3x / (square of X + 4) = 3x / (x + 4 / x)
2. Use the mean inequality to know - 3 / 4
Find the first derivative, calculate the maximum value, you know the range~
If the system of inequalities {x-12m} has no real solution, the range of M is obtained
x-1＜m① x＋1＞2m②
From (1), we get x < m + 1
From 2, we get x ＞ 2m-1
No solution
∴m＋1≤2m-1
∴m≥2
x2m-1
There is no real solution
2m-1>=m+1
m>=2
The square of function f (x) = 3x - 5x + 2. The range of X ∈ [0,2] is
Can it be solved by other methods?
Yes, f (x) = 3x & # 178; - 5x + 2
If f (x) '= 6x-5, Let f (x)' = 0 get x = 5 / 6, then the function is decreasing on [0.5/6], and increasing on (5 / 6,2], then
If the maximum value is f (2) = 4 and the minimum value is f (5 / 6) = 8 / 25, the range is [8 / 25,4]
If the system of inequalities 3-2x about X is greater than or equal to 0, X is greater than or equal to m, and there is a solution, find the value range of M
X
Determine the square-2x-3 range and monotone interval of function y = 3x
Square of x-2x-3 upper right corner
y=3^(x^2-2x-3)
Let m = x ^ 2-2x-3 = (x-1) ^ 2-4 > = - 4
So Y > = 3 ^ (- 4) = 1 / 81, that is, the range is [1 / 81, + infinity]
The axis of symmetry of M is x = 1
x> If M = 1, M is monotonically increasing, then the monotonically increasing interval of the function is [1, + infinite]
X
Y = 3 (x square + 2x) - 3
=3 (x-1 / 3) square-10 / 3
So the range of Y is (- 10 / 3, positive infinity)
Monotone interval: (- infinite, 1 / 3) monotone decreasing
(1 / 3, + infinite) monotone increasing
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