\(\dfrac{x}{4}=\dfrac{y}{7}\)
=>\(\dfrac{x}{20}=\dfrac{y}{35}\)
\(\dfrac{y}{5}=\dfrac{z}{6}\)
=>\(\dfrac{y}{35}=\dfrac{z}{42}\)
Do đó: \(\dfrac{x}{20}=\dfrac{y}{35}=\dfrac{z}{42}=k\)
=>x=20k; y=35k; z=42k
Tính\(\dfrac{3x-4y+5z}{x-2y+5x}\)
\(=\dfrac{3\cdot20k-4\cdot35k+5\cdot42k}{20k-2\cdot35k+5\cdot20k}=\dfrac{13}{5}\)
Chưa chắc đúng nha bạn !!
\[\begin{cases} \frac{x}{4} = \frac{y}{7} \\ \frac{y}{5} = \frac{z}{6} \end{cases}\]
\[x = \frac{4y}{7}\]
\[z = \frac{6y}{5}\]
\[3x - 4y + 5z = 3\left(\frac{4y}{7}\right) - 4y + 5\left(\frac{6y}{5}\right) = \frac{12y}{7} - 4y + \frac{30y}{5} = \frac{12y}{7} - 4y + 6y = \frac{12y}{7} + 2y\]
\[x - 2y + 5x = \frac{4y}{7} - 2y + \frac{30y}{5} = \frac{4y}{7} - 2y + 6y = \frac{4y}{7} + 4y\]
\[\frac{\frac{12y}{7} + 2y}{\frac{4y}{7} + 4y} = \frac{\frac{12y + 14y}{7}}{\frac{4y + 28y}{7}} = \frac{\frac{26y}{7}}{\frac{32y}{7}} = \frac{26y}{32y} = \frac{13}{16}\]
Vậy kết quả cuối cùng là \(\frac{13}{16}\).
