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Anonymous users2024-02-08The process of calculation is correct, but I don't notice the size relationship implied in the question.
x2+y2 ≥2xy
So. a2+2a-3 ≥ 3a2-6a+42a2-8a+7 ≤0
Solution. 2-(root number 2) 2 A 2+ (root number 2) 2, and 2- (root number 2) 2 is greater than 1, so a cannot take 1, so monotonicity should be used.
2xy = 3a^2 - 6a +4
3(a-1)^2 +1
In a monotonically increasing size of 1.
So a = 2-(root number 2) 2 to obtain the minimum value.
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Anonymous users2024-02-072xy=[x+y] -x-squared + y-squared.
2xy={2a-1} squared-{a squared+2a-3}xy=3 2*asquared-3a+2 So when a=1, xy has a minimum value of 1 2 substituted 2xy={2a-1} squared-{a squared+2a-3} left = right so holds.
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Anonymous users2024-02-06x+y=2a-1
x²+y²=a²+2a-3
x+y)²-2xy=a²+2a-3
2a-1)²-2xy=a²+2a-3
xy=(3a -6a+4) 2=[3(a-1) +1] 2x,y are both real numbers.
The equation m -(2a-1)m+=[3(a-1) +1] 2=0 with x and y as the root has a real root.
i.e. [-(2a-1)] 4 [3(a-1) +1] 2 0 solution (4- 2) 2 a (4+ 2) 2 so when a=(4- 2) 2, xy is the smallest.
Minimum = (11-6 2) 4
Note: Your solution does not take into account that "x and y are both real numbers".
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Anonymous users2024-02-05This question is not very difficult, x+y=2a-1 ——
x^2+y^2=a^2+2a-3 ——
2xy = square -
3a^2 - 6a + 4
xy=(3/2)a^2 - 3a + 2
If a=1 can be taken when xy has a minimum value of 1 3, but here 2 * is squared to 0
Get ( 2- 2) 2 a (2+ 2) 2, so a=( 2- 2) 2 has a minimum value.
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