Given x = 3 + 12, y = 3-12, find the value of 3x2-5xy + 3y2

Given x = 3 + 12, y = 3-12, find the value of 3x2-5xy + 3y2

∵ x = 3 + 12, y = 3-12, ∵ x + y = 3, xy = 12, ∵ original formula = 3 (x + y) 2-11, xy = 3 × (3) 2-11 × 12 = 72

Given XY ≠ 1,2x2 + 2006X + 3 = 0,3y2 + 2006y + 2 = 0, try to find the ratio of X and y

First of all, change the two equations into 2x2 + 2006X = - 3 3y2 + 2006y = - 2
Then divide the two formulas and let X / y = a, where a > 0, then 4a2 + 6a-9 = 0
The solution is a = 3 √ 5-3 / 4

Given x > y > 0 and xy = 1, find the minimum value of (x2 + Y2) / (X-Y) and the corresponding value of X, y

(x2 + Y2) / (X-Y) = (x2 + y2-2xy + 2XY) / (X-Y) because xy = 1, so = [(X-Y) ^ 2 + 2] / (X-Y) = (X-Y) + 2 / (X-Y) because x > y > 0, so (X-Y) > 0, so there is an inequality theorem. We know that (X-Y) + 2 / (X-Y) > = 2 root sign 2 under 2 root sign [(X-Y) * 2 / (X-Y)] = 2 root sign 2, and then (X-Y) ^ 2 = 2 conforms to the above