If the real number m, N, x, y satisfies M2 + N2 = a, X2 + y2 = B (a ≠ b), then the maximum value of MX + NY is the best to use the basic inequality That 2 is the square, the answer is the root, AB, not anything else

If the real number m, N, x, y satisfies M2 + N2 = a, X2 + y2 = B (a ≠ b), then the maximum value of MX + NY is the best to use the basic inequality That 2 is the square, the answer is the root, AB, not anything else

Let m = √ a * Sint
Then n ^ 2 = A-A (Sint) ^ 2 = a (cost) ^ 2
Because the cost range is symmetric about the origin
So let n = √ acost
Let x = √ bcosu,
The same as above, y = √ bsinu
mx+ny=√(ab)sintcosu+√(ab)costsinu
=√(ab)(sintcosu+costsinu)
=√(ab)*sin(t+u)
So the maximum value = √ (AB)
I know that 2 is a square, and √ is a root sign. The answer is also a root sign ab

If the ellipse x2 / M2 + Y2 / N2 = 1, m ∈ {1,2,3}, n ∈ {4,5,6,7}, then it can be constructed____ It's a different ellipse

There are three elements in M and four elements in N, and they are different from each other
So 3x4 = 12 different ellipses can be formed

If M2 + 2Mn + 2n2-6n + 9 = 0, find the value of m in N2

m²+2mn+n²+n²-6n+9=0
(m+n)²+(n-3)²=0
∴m+n=0
n-3=0
∴m=-3
n=3
∴m/n²=-3/9=-1/3