When the values of a and B are respectively, the polynomial 4A ^ 2 + B ^ 2 + 4a-6b-8 has the minimum value, and the minimum value can be obtained

When the values of a and B are respectively, the polynomial 4A ^ 2 + B ^ 2 + 4a-6b-8 has the minimum value, and the minimum value can be obtained

Separate part a from part B directly
Original formula = (2a + 1) ^ 2 + (B-3) ^ 2-18
So when a = - 1 / 2, B = 3, the polynomial is the smallest, and the value is - 18
Original formula = (2a + 1) ^ 2 + (B-3) ^ 2
When a = 1 / 2, B = 3, the value of the original formula is the smallest, equal to 0
Given the set a = {X / X & # 178; - 3x-10 ≤ 0} B = {X / M + 1 ≤ x ≤ 2m-1} if B is included in a, find the value range of M
x^2-3x-10
A={x/x²-3x-10≤0}
(x+2)(x-5)
If point P is the golden section point of segment AB, what is the ratio of the shorter segment PA to the longer segment Pb?
PA:PB=0.618
The value of P point is 0.618 (golden section point), the short-term PA is 0.382, and the long-term Pb is 0.618. The PA / PD ratio is 382 / 618
When the values of a and B are what, the polynomial A2 + 6A + b2-10b + 40 has the minimum value? And find the minimum
A2 + 6A + b2-10b + 40 = A2 + 6A + 9 + b2-10b + 25 + 6 = (a + 3) 2 + (B-5) 2 + 6 when a + 3 = 0 and B-5 = 0, that is, when a = - 3 and B = 5, the minimum value of (a + 3) 2 + (B-5) 2 + 6 is 6. So when a = - 3 and B = 5, the minimum value of polynomial A2 + 6A + b2-10b + 40 is 6
Let a = {x | X & # 178; - MX + 6 = 0} and B = {x | X & # 178; - 2x + n = 0}. Given a ∩ B = {3}, find the value of M and n
A∩B=﹛3﹜
X = 3 is the solution of X & # 178; - MX + 6 = 0
9-3m+6=0,m=5
X = 3 is the solution of X & # 178; - 2x + n = 0
9-6+n=0,n=-3
If C is the golden section point of line AB, take point D on the longer line AC so that CD = BC, and prove that point D is the golden section point of ca
More details, thank you
Let AB = 1, then AC = (√ 5-1) / 2, BC = (3 - √ 5) / 2, then CD = (3 - √ 5) / 2, ad = √ 5-2
Then ac * ad = (√ 5-1) / 2 * (√ 5-2) = (7-3 √ 5) / 2;
CD=(3-√5)^2/4=(7-3√5)/2=AC*CD
So D is the golden section of ca
When a, B what value, polynomial A & sup2; - B & sup2; - 4A + 6B + 18 has the minimum value? And find out the minimum value
This formula has no minimum value, because the coefficient of B ^ 2 is - 1. If a does not take infinity and B takes infinity, then the formula can take negative infinity
If the original formula is a & sup2; + B & sup2; - 4A + 6B + 18
Original formula = (a ^ 2-4a + 4) + (b ^ 2 + 6B + 9) + 5
Then a is 2, B = - 3, and the minimum value is 5
Anyway, the problem of taking extreme value, if you haven't learned derivative, it's all formula, and then the things in the square must be greater than or equal to 0, you can judge
The minimum order 23 A and B are combined into a square term
(a-2)^2+(3-b)^2+23
A-2 = 0.3-b = 0 is the smallest, that is, a = 2, B = 3
Wrong? It's not worth it
Given the complete set u = R and the set M = {x ᦉ 178; - 4 ≤ 0}, then M complement = ()
∵ u = R, M = {x ᦉ 178; - 4 ≤ 0} = {X - 2 ≤ x ≤ 2}, the complement of ∵ M = {x ﹤ 2 or X ﹥ 2}
But I don't understand how {X - 2 ≤ x ≤ 2} came about,
Solving inequality X & # 178; - 4 ≤ 0
That is, (x + 2) (X-2) ≤ 0
The solution is - 2 ≤ x ≤ 2
Complete set u = R
Therefore, the complement of M = {x ﹤ 2 or X ﹥ 2}
If C is the golden section point of line AB, take a point D on the longer line AC so that CD = BC, and prove that point D is the golden section point of ca
Sheila
Let AB = 1, then AC = (√ 5-1) / 2, BC = (3 - √ 5) / 2, then CD = (3 - √ 5) / 2, ad = √ 5-2
Then ac * ad = (√ 5-1) / 2 * (√ 5-2) = (7-3 √ 5) / 2;
CD=(3-√5)^2/4=(7-3√5)/2=AC*CD
So D is the golden section of ca
When a, B what value, polynomial A's square + B's square + 2a-4b + 16 have minimum value? Try to find this minimum value
When the square of (a + 1) + the square of (b-2) + 11 A = - 1 and B = 2, the minimum is 11
a²+2a+1+b²-4b+4+11
=（a+1）²+（b-2）²+11
When B = - 11, a is the minimum
(a + 1) (a + 1) + (b-2) (b-2) + 11 =? The minimum value is 11
When a = - 1, B = 2, the minimum is taken
=a^2+2a+1+b^2-4b+4+11
=(a+1)^2+(b-2)^2+11
From = (a + 1) ^ 2 ≥ 0 (b-2) ^ 2 ≥ 0, we get
=(a+1)^2+(b-2)^2+11≥11
So the minimum is 11