tìm x, y, z biết: $\frac{x-1}{2}$ = $\frac{y+3}{4}$ = $\frac{z-5}{6}$ và 5z+3x-4y=50

Ta có:

$\dfrac{x -1}{2} = \dfrac{y+3}{4} = \dfrac{z – 5}{6}$

$\to \dfrac{3(x -1)}{2.3} = \dfrac{4(y+3)}{4.4} = \dfrac{5(z – 5)}{6.5}$

$\to \dfrac{3x -3}{6} = \dfrac{4y+12}{16} = \dfrac{5z – 25}{30}$

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

$\dfrac{5z – 25}{30} = \dfrac{3x -3}{6} = \dfrac{4y+12}{16} = \dfrac{5z – 25 + 3x – 3 – 4y – 12}{30 + 6 – 16} = \dfrac{50- 25- 3 – 12}{20} = \dfrac{1}{2}$

$+)\quad \dfrac{x -1}{2} = \dfrac{1}{2} \Leftrightarrow x – 1 = 1\Leftrightarrow x =2$

$+)\quad \dfrac{y +3}{4}=\dfrac{1}{2}\Leftrightarrow y + 3 = 8 \Leftrightarrow y = 5$

$+)\quad \dfrac{z -5}{6}=\dfrac{1}{2}\Leftrightarrow z – 5 = 3\Leftrightarrow z = 8$

Vậy $(x;y;z) = (2;5;8)$