Ta có: \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21};5x+y-2z=28\)
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)
\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42};5x+y-2z\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
Suy ra: \(\frac{5x}{50}=2\Rightarrow x=2.50:5=10\)
\(\frac{y}{6}=2\Rightarrow y=2.6=12\)
\(\frac{2z}{42}=2\Rightarrow z=2.42:2=42\)
Vậy \(x=20;y=12;z=42\)
b) 3x = 2y; 7y = 5z; x - y + z =32
=> \(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\)
=> \(\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\)
=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{x-y+z}{16}\)
=> \(\frac{x}{10}=2\Rightarrow x=20\)
=> \(\frac{y}{15}=2\Rightarrow y=30\)
=> \(\frac{z}{21}=2\Rightarrow z=42\)
d,
ta có:\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x.y.z}{2.3.5}=\frac{810}{30}=27\)
\(=>\frac{x}{2}=27=>x=2.27=54\)
\(=>\)\(\frac{y}{3}=277=>y=3.27=81\)
\(\frac{z}{5}=27=>z=5.27=135\)
vậy:x=54,y=81,z=135