Có: \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
<=> \(\dfrac{3\left(x-1\right)}{3.2}=\dfrac{4\left(y+3\right)}{4.4}=\dfrac{5\left(z-5\right)}{6.5}\)
<=> \(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)
mà 5z-3x-4y=50
ADTCDTSBN ta có:\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{5z-25-\left(4y+12\right)-\left(3x-3\right)}{30-16-6}=\dfrac{5z-25-4y-12-3x+3}{8}=\dfrac{\left(5z-4y-3x\right)-\left(25+12-3\right)}{8}=\dfrac{50-34}{8}=2\)
Do đó: \(\dfrac{3x-3}{6}=2\) <=> \(\dfrac{x-1}{2}=2\) <=> x-1=4 <=> x=5
\(\dfrac{4y+12}{16}=2\) <=> \(\dfrac{y+3}{4}=2\) <=> y+3=8<=> y=5
\(\dfrac{5z-25}{30}=2\) <=> \(\dfrac{z-5}{6}=2\) <=> z-5=12 <=> z=17
Vậy x=5 , y=5 , z=17
