Given that 2A ^ 2 - 8ab + 17b ^ 2 - 16A - 4B + 68 is greater than or equal to 0, find the value of a and B

Given that 2A ^ 2 - 8ab + 17b ^ 2 - 16A - 4B + 68 is greater than or equal to 0, find the value of a and B

Should be less than or equal to 0?
2a*a-8ab+17b*b-16a-4b+68
=(a-4b) square + (A-8) square + (b-2) square=
Given the complete set u = R, a = {the square of X / x-3x + 2 ＜ = 0} B = {the square of X / x-2ax + a ＜ = 0, a belongs to R} if a is combined with B, the value range of a is obtained
A = {the square of X / X - 3x + 2 ＜ = 0} = [1,2]
B = {the square of X / X - 2aX + a ＜ = 0, a belongs to R} = {a}
If a and B = a, → a ∈ [1,2]
The value range of a [1,2]
The 5cm line segment is golden section, how many cm shorter?
The long one is 5 * 0.618 = 3.09
The short is 5-3.09 = 1.91
The long one is (√ 5-1) / 2 * 5,
The short one is 5 - (√ 5-1) / 2 * 5 = (15-5 √ 5) / 2cm
Length of short segment: 5 * (1-0.618) = 1.91 (CM).
golden section
A line segment is divided into two parts so that the ratio of one part to the whole length is equal to the ratio of the other part to this part. The first three digits of the ratio are the irrational number 618. Because the shape designed according to this proportion is very beautiful, it is called golden section, also known as Sino foreign ratio. This is a very interesting number. We take 0.618 as an approximation. Through simple calculation, we can find that:
1/0.618=1.618
(1-0.618)/0.618=0.618
The expansion of this number
golden section
A line segment is divided into two parts so that the ratio of one part to the whole length is equal to the ratio of the other part to this part. The ratio is an irrational number, and the approximate value of the first three digits is 0.618. According to this, it is also called golden section design. This is a very interesting number. We take 0.618 as an approximation. Through simple calculation, we can find that:
1/0.618=1.618
(1-0.618)/0.618=0.618
The function of this value is not only reflected in the art fields such as painting, sculpture, music and architecture, but also plays an important role in management and engineering design.
5cm is about 3.09 in length. 91. Put it away
It is known that f (x) is an even function defined on R, and if x ≤ 0, f (x) = log4 (- x + 1)
If the inequality f (2 ^ t ● a) ≤ t holds for t ∈ [1, ∞), the value range of real number a is obtained
Because f (x) is an even function, and if x ≤ 0, f (x) = log4 (- x + 1), then f (x) is: F (x) = log4 (| x | + 1) on R. thus, the inequality f [(2 ^ t) ● a] ≤ t is log4 (| (2 ^ t) ● a | + 1) ≤ t
The simplification is: (2 ^ t) a + 1 ≤ 4 ^ t
(2^t)a＋1≤[2^t]²
Let m = 2 ^ t, because t ≥ 1, then m ≥ 2,
That is, the inequality Ma + 1 ≤ M & # 178 holds for all m ≥ 2
a≤m－(1/m)
Considering that M - (1 / M) is increasing when m ≥ 2, if a ≤ m - (1 / M) is constant for all m ≥ 2, then:
A ≤ [M - (1 / M)]
a≤3/2
f（2^t●a）=log4(-2^t●a+1)≤t
t=log4(4^t)
log4(-2^t●a+1)≤log4(4^t)
-2^t●a+1≤4^t
a> = (1-4 ^ t) / 2 ^ t is constant for t ∈ [1, ∞), that is, a is greater than or equal to the maximum value of (1-4 ^ t) / 2 ^ t
And (1-4 ^ t) / 2 ^ t = 1 / (2 ^ t) - 2 ^ t
When t is larger, 2 ^ t is larger, then
f（2^t●a）=log4(-2^t●a+1)≤t
t=log4(4^t)
log4(-2^t●a+1)≤log4(4^t)
-2^t●a+1≤4^t
a> = (1-4 ^ t) / 2 ^ t is constant for t ∈ [1, ∞), that is, a is greater than or equal to the maximum value of (1-4 ^ t) / 2 ^ t
^ 2 (t-1) / T-1
When t is larger, 2 ^ t is larger, then 1 / (2 ^ t) and - 2 ^ t are smaller
So when t = 1, the maximum value of (1-4 ^ t) / 2 ^ t is - 3 / 2
A > = - 3 / 2
Given that the complete set is r, a = {x | x 2-3x + 2 ≤ 0}, B = {x 2-2a x = a 2 ≤ 0}, and B ∪ a = a, find the value range of a (a ∈ R)
A [1,2]
B x=a
So the value of a is [1,2]
A = [1,2] should be a closed interval
B={a}
B ∪ a = a, that is, a belongs to a = [1,2]
So the value range of a is [1,2], which is also a closed interval
Follow these steps to draw a picture
Draw a square ABCD
Take the midpoint e of AB to connect CE
Extend AB to point F so that EF = CE
Draw a square with BF as the edge bFGH BF is less than ab
So is h the golden section of edge AB? Why
Yes, he satisfies the golden ratio (root (5) - 1) / (2)
It is known that f (x) is an even function defined on R, and X ≤ 0, f (x) = log4 (- x + 1) f (A-1) - f (3-A)
If f (A-1) - f (3-A) A-3 is a > 3 when A-1
Given a = {x ∈ R | X & # 178; - 3x + 4 = 0}, B = {x ∈ R | (x + 1) (X & | 178; + 3x-4 = 0), a is really contained in P and contained in B, find the set p? And write the process
A={x∈R|x²-3x+4=0}
B={x∈R|(x+1)(x²+3x-4)=0}
Because the discriminant of X & # 178; - 3x + 4 = 0 is Δ = (- 3) ^ 2-4 * 1 * 4 = - 7 < 0
So x & # 178; - 3x + 4 is always greater than 0
Then a = an empty set
(x+1)(x²+3x-4)=0
X + 1 = 0 or X & # 178; + 3x-4 = 0
So x = - 1 or x = 1 or x = - 4
B={-4,-1,1}
A is really contained in P and contained in B
So p = {- 4} or {- 1} or {1} or {- 4, - 1} or {- 4,1} or {- 1,1} or {- 4, - 1,1}
If you don't understand, please hi me, I wish you a happy study!
A={x∈R|x²-3x+4=0}=Φ,B={x∈R|（x+1）(x²+3x-4=0)={-4，-1,1}，
And a is really contained in P and B, so p has the following possibilities:
{-4}，{-1}，{1}，{-4，-1}，{-4,1}，{-1,1}，{-4，-1,1}
Mathematical golden section
(1) AB = 2, AC = radical 5-1, please prove AC & sup2; = AB * BC
(2) AB = 2, point C is the golden section point of AB, point D is on AB, and AD & sup2; = BD * AB, find the ratio of CD to AC (as shown in the figure)————
A C D B
(1) AC & sup2; = (√ 5-1) & sup2; = 6-2 √ 5ab * BC = AB * (AB-AC) = 2 (2 - (√ 5-1)) = 6-2 √ 5, so AC & sup2; = AB * BC (2) because C and D are the golden section of AB, AC = BD = √ 5-1cd = ab-ac-bd = 2-2 (√ 5-1) = 4-2 √ 5CD / AC = (4-2 √ 5) / (√ 5-1) = (4-2 √ 5) (√ 5 + 1) / 4 =
Let f (x) be an odd function defined on R, and if x ∈ [0, + ∞), f (x) = x (1 + 3 √ x), find the analytic expression of F (x)
3 √ x is the cube root of X~
Because odd functions are
So it's easy to get f (- x) = - f (x)
So f (- x) = - x (1 + 3 √ x)
Because x ∈ [0, + ∞)
So - x ∈ (- ∞, 0)
So think of - x as X
f(x)=x(1+3√-x)
therefore
f(x)=x(1+3√x),X∈[0,+∞）
f(x)=x(1+3√-x),x∈（-∞,0）