Em mời các chuyên gia ạ! Question Em mời các chuyên gia ạ! in progress 0 Môn Toán Jezebel 5 years 2021-05-17T15:38:03+00:00 2021-05-17T15:38:03+00:00 2 Answers 14 views 0
Answers ( )
Chúc bạn học tốt :3 Mình xin CTLHN+TIM Ạ :3
Đáp án:
$\min D = -1 \Leftrightarrow x = 3y$
$\max D = \dfrac97 \Leftrightarrow y = 3x$
Giải thích các bước giải:
$\quad D =\dfrac{5y^2 – 3xy}{x^2 – 3xy + 4y^2}$
$+)\quad \min$
$\quad D + 1 =\dfrac{5y^2 – 3xy}{x^2 – 3xy + 4y^2} + 1$
$\to D + 1 = \dfrac{5y^2 – 3xy + x^2 – 3xy + 4y^2}{x^2 – 3xy + 4y^2}$
$\to D + 1 = \dfrac{9y^2 – 6xy + x^2}{x^2 – 3xy + 4y^2}$
$\to D + 1= \dfrac{(3y – x)^2}{x^2 – 3xy + 4y^2}$
$\to D + 1\geqslant 0$
$\to D \geqslant -1$
Dấu $=$ xảy ra $\Leftrightarrow 3y – x = 0 \Leftrightarrow x = 3y$
$+)\quad \max$
$\quad D – \dfrac97 = \dfrac{5y^2 – 3xy}{x^2 – 3xy + 4y^2} – \dfrac97$
$\to D -\dfrac97 =\dfrac{7(5y^2- 3xy) – 9(x^2 – 3xy + 4y^2)}{7(x^2 – 3xy + 4y^2)}$
$\to D -\dfrac97 =\dfrac{-9x^2 + 6xy – y^2}{7(x^2 – 3xy + 4y^2)}$
$\to D – \dfrac97 = -\dfrac{(y – 3x)^2}{7(x^2 – 3xy + 4y^2)}$
$\to D -\dfrac97 \leqslant 0$
$\to D \leqslant \dfrac97$
Dấu $=$ xảy ra $\Leftrightarrow y – 3x = 0 \Leftrightarrow y = 3x$
Vậy $\min D = -1 \Leftrightarrow x = 3y$
$\max D = \dfrac97 \Leftrightarrow y = 3x$