Given a ^ 2 + 2B + B ^ 2-4b + 5 = 0, find the value of 2A ^ 2 + 4b-3

Given a ^ 2 + 2B + B ^ 2-4b + 5 = 0, find the value of 2A ^ 2 + 4b-3

Given that rational numbers a and B satisfy a (a + 1) - (a 2 + 2b) = 1, find the value of a 2-4ab + 4B 2-2a + 4B
If √ a-2b + 5 and (a + B-1) & sup2; are opposite to each other, find the arithmetic square root of b-a
Given the quadratic power of a - 2A + the quadratic power of B + 4B + 5 = 0, find the value of (a + b) to the power of 2012
The 1900 is for you
The original formula can be arranged as follows:
a^2-2a+1 + b^2+4b+4 =0
(a-1)^2 + (b+2)^2 =0
So a = 1, B = - 2
(a+b)^2012=(-1)^2012=1
If a = A2 + 5b2-4ab + 2B + 100, find the minimum value of A
The minimum value of a = a2-4ab + 4B2 + B2 + 2B + + 1 + 99 = (a-2b) 2 + (B + 1) 2 + 99, ∵ (a-2b) 2 ≥ 0, (B + 1) 2 ≥ 0, ∵ a ≥ 99, ∵ A is 99
Let a, B ∈ R, A2 + 2B2 = 6, then the minimum value of a + B is______ .
A2 + 2B2 = 6, can be changed to A26 + B23 = 1, so let a = 6cos θ, B = 3sin θ, then a + B = 6cos θ + 3sin θ = 3 (63cos θ + 33sin θ) & nbsp; θ∈ [0, 2 π] let Tan α = 2, then a + B = 3sin (θ + α) ≥ - 3 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; θ∈ [0, 2 π], then the minimum value of a + B is - 3
Let a, B ∈ R, A2 + 2B2 = 6, then the minimum value of a + B is? Why use trigonometric function
Let a, B ∈ R, A2 + 2B2 = 6, then the minimum value of a + B is______ .
Why use trigonometric function?
The way to use trigonometric function is to treat it as a quadratic curve and solve it by parametric equation, which has only one parameter and is much more convenient
For example, if this problem is an ellipse, let a = √ 6cost, B = √ 3sint
a+b=√6cost+√3sint=3sin(t+p),tanp=√6/√3=√2
So the minimum is - 3 and the maximum is 3
Use triangle to change element.
Let a, B ∈ R, A2 + 2B2 = 6, then the minimum value of a + B is______ .
A2 + 2B2 = 6, can be changed to A26 + B23 = 1, so let a = 6cos θ, B = 3sin θ, then a + B = 6cos θ + 3sin θ = 3 (63cos θ + 33sin θ) & nbsp; θ∈ [0, 2 π] let Tan α = 2, then a + B = 3sin (θ + α) ≥ - 3 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; θ∈ [0, 2 π], then the minimum value of a + B is - 3
How to calculate the derivative of F (x) = ln (3-2x ^ 2) and what is the monotone interval
f（x）=ln（3-2x^2）
Domain 3-2x & # 178; > 0
2x²
=3
f^=-4x / 3-2x^2
3-2x ^ 2
Therefore, when x > 0, f ^ > 0 is constant, i.e. the increasing interval of F can be known
X
Let a and B be real numbers, find the minimum value of A2 + 2Ab + 2B2 + 2b2-4b-5, and find the value of a and B at this time
a2+2ab+2b2-4b-5
=a²+2ab+b²+b²-4b+4-9
=(a+b)²+(b-2)²-9
When a + B = 0
When B-2 = 0, there is a minimum value of - 9
Here a = - 2, B = 2