Question

Tìm x,y,z:     \(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)     và 5z-3x-4y=50

Answer 1

\(\frac{x-1}{2}=\frac{y=3}{4}=\frac{z-5}{6}\Rightarrow\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)

Áp dụng tính chất dãy TSBN, ta có:

\(\frac{\left(5z-25\right)-\left(3x-3\right)-\left(4x+12\right)}{30-6-16}\)

= \(\frac{\left(5z-3x-4y\right)-\left(25-3+12\right)}{8}\)

= \(\frac{50-34}{8}\)

= \(\frac{16}{8}\)

= 2

Suy ra:  \(\frac{3x-3}{6}=2\Rightarrow\frac{x-1}{2}=2\Rightarrow x=5\)

\(\frac{4y+12}{6}=2\Rightarrow\frac{y+3}{4}=2\Rightarrow y=5\)

\(\frac{5z-25}{30}=2\Rightarrow\frac{z-5}{6}=2\Rightarrow z=17\)

Vậy: x = 5

         y = 5

         z = 1

Answer 2

kik cho mik