Cho 4x=5y; 8z=6y và 2x-|3y - z|=-19. Tìm x, y, z. HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEE

$4x=5y\\↔y=\dfrac{4x}{5}\\8z=6y\\↔z=\dfrac{3y}{4}$

Thay $y=\dfrac{4x}{5}$ vào biểu thức $z=\dfrac{3y}{4}$

$z=\dfrac{3.\dfrac{4x}{5}}{4}=\dfrac{12x}{5}.\dfrac{1}{4}=\dfrac{3x}{5}$

Thay $y=\dfrac{4x}{5},z=\dfrac{3x}{5}$ vào biểu thức $2x-|3y-z|=-19$

$↔\left|3.\dfrac{4x}{5}-\dfrac{3x}{5}\right|=2x+19\\↔\left|\dfrac{12x}{5}-\dfrac{3x}{5}\right|=2x+19\\↔\left|\dfrac{9x}{5}\right|=2x+19$

Ta có: $\left|\dfrac{9x}{5}\right|=\begin{cases}\dfrac{9x}{5}\,\,nếu\,\,\dfrac{9x}{5}\ge 0\,\,hay\,\,x\ge 0\\-\dfrac{9x}{5}\,\,nếu\,\,\dfrac{9x}{5}<0\,\,hay\,\,x<0\end{cases}$

TH1: $x\ge 0$

$→\dfrac{9x}{5}=2x+19\\↔-\dfrac{x}{5}=19\\↔x=-95(KTM)$

$→y,z$ không có giá trị thỏa mãn

TH2: $x<0$

$→-\dfrac{9x}{5}=2x+19\\↔-\dfrac{19x}{5}=19\\↔19x=-95\\↔x=-5(TM)\\→y=\dfrac{4.(-5)}{5}=-4\\→z=\dfrac{3.(-4)}{4}=-3$

Vậy $(x;y;z)=(-5;-4;-3)$