Đáp án:
Ta có :
`x^2 + y^2 = x + y <=> x^2 - x + y^2 - y = 0 <=> x^2 - x + 1/4 + y^2 - y + 1/4 = 1/2`
`<=> (x - 1/2)^2 + (y - 1/2)^2 = 1/2`
Lúc đó : `x + 2y = 1.(x - 1/2) + 2.(y - 1/2) + 3/2` , Áp dụng BĐT `bu-nhi-a`
`-> (1.(x - 1/2) + 2.(y - 1/2))^2 <= (1^2 + 2^2)((x - 1/2)^2 + (y - 1/2)^2) = 5 . 1/2 = 5/2`
`-> (1.(x - 1/2) + 2.(y - 1/2))^2 <= 5/2`
`-> -\sqrt{5/2} <= x - 1/2 + 2(y - 1/2) <= \sqrt{5/2}`
`-> 3/2 - \sqrt{5/2} <= x + 2y <= \sqrt{5/2} + 3/2`
Vậy $Min_{F}$ là `3/2 - \sqrt{5/2} <=> x = -\sqrt{1/10} + 1/2 ; y = -\sqrt{2/5} + 1/2`
`Max_{F} = \sqrt{5/2} + 3/2 <=> x = \sqrt{1/10} + 1/2 ; y = \sqrt{2/5} + 1/2`
Giải thích các bước giải:
