\(\frac{3x}{4}=\frac{y}{2}=\frac{3z}{5}\)
\(\Rightarrow\frac{x}{\frac{4}{3}}=\frac{y}{2}=\frac{z}{\frac{5}{3}}=\frac{x+y+z}{5}=\frac{y-z}{\frac{1}{3}}=\frac{15}{\frac{1}{3}}=45\)
\(\Rightarrow x+y+z=45.5=225\)
PS: Bài hồi nãy làm nhầm nhé
\(\frac{3x}{4}=\frac{y}{2}=\frac{3z}{5}=\frac{x}{\frac{4}{3}}=\frac{y}{\frac{2}{1}}=\frac{z}{\frac{5}{3}}\)
Áp dụng TC DTSBN ta có :
\(\frac{x}{\frac{4}{3}}=\frac{y}{\frac{2}{1}}=\frac{z}{\frac{5}{3}}=\frac{y-z}{\frac{2}{1}-\frac{5}{3}}=\frac{15}{\frac{1}{3}}=45\)
\(\Rightarrow\frac{x}{\frac{4}{3}}=45\Rightarrow x=45.\frac{4}{3}=60\)
\(\Rightarrow\frac{y}{\frac{2}{1}}=45\Rightarrow y=45.2=90\)
\(\Rightarrow\frac{z}{\frac{5}{3}}=45\Rightarrow z=45.\frac{5}{3}=75\)
\(\Rightarrow x+y+z=60+90+75=225\)
\(\frac{3x}{4}=\frac{y}{2}=\frac{3z}{5}\)
\(\Rightarrow\frac{x}{\frac{4}{3}}=\frac{y}{2}=\frac{z}{\frac{5}{3}}=\frac{x+y+z}{5}=\frac{x-z}{-\frac{1}{3}}=\frac{15}{-\frac{1}{3}}=-45\)
\(\Rightarrow x+y+z=-45.5=-225\)