Question

Cho \(x^2+4y^2=1\). Tìm Min, Max: \(S=x+y\)

Answer 1

Ta có hai hằng đẳng thức:

\((x - y)^2 = x^2 + y^2 - 2xy <=> 4(x - y)^2 = 4x^2 + 4y^2- 8xy\) (1*)

\((x+4y)^2 = x^2 + 16y^2 + 8xy\) (2*)

Lấy (1*) + (2*): \(4(x-y)^2 + (x+4y)^2 = 5x^2 + 20y^2\)

\(= 5(x^2 + 4y^2) = 5\)

Có: \(4(x-y)^2 ≤ 4(x-y)^2 + (x+4y)^2 = 5\)

\(=> (x-y)^2 ≤ 5/4 => x-y ≤\) \(\dfrac{\sqrt{5}}{2}\)

Dấu "=" khi \(x+4y=0\) và \(x-y=\dfrac{\sqrt{5}}{2}\Rightarrow x=\dfrac{2\sqrt{5}}{5};y=\dfrac{-\sqrt{5}}{10}\)

Vậy \(MAX\left(x-y\right)=\dfrac{\sqrt{5}}{2}\) xảy ra khi \(x=\dfrac{2\sqrt{5}}{5};y=\dfrac{-\sqrt{5}}{10}\)