Toán tìm gtnn p=xy/x^2+y^2 +(1/x+1/y).căn 2(x^2+y^2) 14/09/2021 By Sadie tìm gtnn p=xy/x^2+y^2 +(1/x+1/y).căn 2(x^2+y^2)
Đáp án: $P\ge\dfrac92$ Giải thích các bước giải: Ta có: $P=\dfrac{xy}{x^2+y^2}+(\dfrac1x+\dfrac1y)\cdot\sqrt{2(x^2+y^2)}$ $\to P\ge \dfrac{xy}{x^2+y^2}+(\dfrac1x+\dfrac1y)\cdot\sqrt{(x+y)^2}$ $\to P\ge \dfrac{xy}{x^2+y^2}+(\dfrac1x+\dfrac1y)\cdot(x+y)$ $\to P\ge \dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}}+(\dfrac{x}{y}+\dfrac{y}{x})+2$ $\to P\ge \dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}}+\dfrac14(\dfrac{x}{y}+\dfrac{y}{x})+\dfrac34(\dfrac{x}{y}+\dfrac{y}{x})+2$ $\to P\ge 2\sqrt{\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}}\cdot \dfrac14(\dfrac{x}{y}+\dfrac{y}{x})}+\dfrac34\cdot 2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}+2$ $\to P\ge \dfrac92$ Dấu = xảy ra khi $x=y$ Trả lời
Đáp án: $P\ge\dfrac92$
Giải thích các bước giải:
Ta có:
$P=\dfrac{xy}{x^2+y^2}+(\dfrac1x+\dfrac1y)\cdot\sqrt{2(x^2+y^2)}$
$\to P\ge \dfrac{xy}{x^2+y^2}+(\dfrac1x+\dfrac1y)\cdot\sqrt{(x+y)^2}$
$\to P\ge \dfrac{xy}{x^2+y^2}+(\dfrac1x+\dfrac1y)\cdot(x+y)$
$\to P\ge \dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}}+(\dfrac{x}{y}+\dfrac{y}{x})+2$
$\to P\ge \dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}}+\dfrac14(\dfrac{x}{y}+\dfrac{y}{x})+\dfrac34(\dfrac{x}{y}+\dfrac{y}{x})+2$
$\to P\ge 2\sqrt{\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{x}}\cdot \dfrac14(\dfrac{x}{y}+\dfrac{y}{x})}+\dfrac34\cdot 2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}+2$
$\to P\ge \dfrac92$
Dấu = xảy ra khi $x=y$