D=|x/5−y/7|+|2z−3x|+|xy+yz−500+zx∣ Tìm min D

Đáp án: $Min_D=0$

Giải thích các bước giải:

Ta có:

$|\dfrac{x}{5}-\dfrac{y}{7}|\ge 0,\quad\forall x,y$

$|2z-3x|\ge 0,\quad\forall x,z$

$|xy+yz-500+zx|\ge 0,\quad\forall x,y,z$

$\to D=|\dfrac{x}{5}-\dfrac{y}{7}|+|2z-3x|+|xy+yz-500+zx|\ge 0$

$\to Min_D=0$

Khi đó:

$\begin{cases}\dfrac{x}{5}-\dfrac{y}{7}=0\\2z-3x=0\\ xy+yz-500+zx=0\end{cases}$

$\to \begin{cases}\dfrac{x}{5}=\dfrac{y}{7}\\2z=3x\\ xy+yz+zx=500\end{cases}$

$\to \begin{cases}y=\dfrac75x\\z=\dfrac32x\\ x\cdot \dfrac75x+\dfrac75x\cdot \dfrac32x+\dfrac32x\cdot x=500\end{cases}$

$\to \begin{cases}y=\dfrac75x\\z=\dfrac32x\\5x^2=500\end{cases}$

$\to \begin{cases}y=\dfrac75x\\z=\dfrac32x\\x^2=100\end{cases}$

$\to \begin{cases}y=\dfrac75x\\z=\dfrac32x\\x=\pm10\end{cases}$

$\to (x,y,z)\in\{(10,14, 6),(-10,-14, -6)\}$