Given that the point m (A.B) is on the straight line x + y = 1, then √ the minimum value of a ^ 2 + B ^ 2

Y=-X+1
a^2+b^2=a^2+(-a+1)^2=2a^2-2a+1
The minimum value of √ a ^ 2 + B ^ 2 is the minimum value of 2A ^ 2-2a + 1 and must be greater than or equal to 0
Then when a = 1 / 2, there is a minimum value √ 2 / 2

It is known that X and Y belong to (0, + ∞), 2 ^ (x-3) = (1 / 2) ^ y. if the minimum value of 1 / x + m / Y (M > 0) is 3, then M is equal to?

The (x-3) power of 2 is equal to the Y power of (1 / 2)
2^(x-3)=2^(-y)
We get x-3 = - y, that is, x + y = 3
Because XY ∈ (0. ∞), there is 0

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