x > y > 0 thoar max xy = 2 x^2 + y^2 / x – y >=4 02/11/2021 Bởi Kaylee x > y > 0 thoar max xy = 2 x^2 + y^2 / x – y >=4
Ta có : $\dfrac{x^2+y^2}{x-y} $ $ = (x-y)^2+2xy}{x-y} $ $ = \dfrac{4}{x-y}+(x-y) ≥ 4$ Dấu “=” xảy ra $⇔x=y=\sqrt[]{2}$ Bình luận
$\frac{x^{2}+y^{2}}{x-y}$ = $\frac{(x-y)^{2}+2xy}{x-y}$ = $(x-y)$ + $\frac{4}{x-y}$ ≥ 2$\sqrt{(x-y)\frac{4}{x-y}}$ = 4 Bình luận
Ta có :
$\dfrac{x^2+y^2}{x-y} $
$ = (x-y)^2+2xy}{x-y} $
$ = \dfrac{4}{x-y}+(x-y) ≥ 4$
Dấu “=” xảy ra $⇔x=y=\sqrt[]{2}$
$\frac{x^{2}+y^{2}}{x-y}$ = $\frac{(x-y)^{2}+2xy}{x-y}$ = $(x-y)$ + $\frac{4}{x-y}$ ≥ 2$\sqrt{(x-y)\frac{4}{x-y}}$ = 4