Đáp án:
$M_{max}=\dfrac{9}{2}$ khi $(x;y)=(1;2)$
Giải thích các bước giải:
Do $1 \leq x<y \leq 2⇒1<\dfrac{y}{x} \leq 2$
Ta có:
$M=2+\dfrac{x}{y}+\dfrac{y}{x}$
Đặt $\dfrac{y}{x}=t⇒1<t \leq 2$
$⇒M=2+t+\dfrac{1}{t}=\dfrac{9}{2}+t+\dfrac{1}{t}-\dfrac{5}{2}$
$⇒M=\dfrac{9}{2}+\dfrac{2t^2-5t+2}{2t}$
$⇒M=\dfrac{9}{2}+\dfrac{(2t-1)(t-2)}{2t}$
Do $1<t \leq 2$$⇒\begin{cases}2t-1>0\\t-2 \leq 0 \end{cases}$
$⇒(2t-1)(t-2) \leq 0$
$⇒\dfrac{(2t-1)(t-2)}{2t} \leq 0$
$⇒M=\dfrac{9}{2}+\dfrac{(2t-1)(t-2)}{2t} \leq \dfrac{9}{2}$
Vậy $M_{max}=\dfrac{9}{2}$ khi $t=2⇔(x;y)=(1;2)$
