Có \(\frac{x}{5}=\frac{y}{6};\frac{y}{8}=\frac{z}{7}\)\(\Rightarrow\frac{x}{40}=\frac{y}{48}=\frac{z}{42}=\frac{x+y-z}{40+48-42}=\frac{138}{46}=3\)
Vậy \(\frac{x}{40}=3\Rightarrow x=120\)
\(\frac{y}{48}=3\Rightarrow y=144\)
\(\frac{z}{42}=3\Rightarrow z=126\)
Ta có: \(\frac{x}{5}\)= \(\frac{y}{6}\)=> \(\frac{x}{20}\)= \(\frac{y}{24}\)
\(\frac{y}{8}\)= \(\frac{z}{7}\)=> \(\frac{y}{24}\)= \(\frac{z}{21}\)
=> \(\frac{x}{20}\)= \(\frac{y}{24}\)= \(\frac{z}{21}\)
Áp dụng tính chất dãy tỉ số = nhau
Ta có: \(\frac{x}{20}\)= \(\frac{y}{24}\)= \(\frac{z}{21}\)= \(\frac{x+y-z}{20+24-21}\)= \(\frac{138}{23}\)= 6
Vậy x = 120
y = 144
z = 126
