Đat 2x/3=3y/4=4z/5=k(k khác 0). Ta co x=3/2.k ; y=4/3k ; z =5/4k. => x+y+z=3/2k+4/3k+5/4k=k.(49/12)=49. =>k=12 =>x=18;y=16 ;z=15
a.=>12x/18=12y/16=12z/15
áp dụng tính chất dãy tỉ số bằng nhau,ta có:
12x/18=12y/16=12z/15=>12x+12y+12z/18+16+15=12*(x+y+z)/49=12*49/49
=>12x/18=12=>x=18
=>12y/16=12=>y=16
=>12z/15=15=>z=15
a) Đặt \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=k\Rightarrow\hept{\begin{cases}2x=3k\\3y=4k\\4z=5k\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3k}{2}\\y=\frac{4k}{3}\\z=\frac{5k}{4}\end{cases}}}\)
\(\Rightarrow x+y+z=\frac{3k}{2}+\frac{4k}{3}+\frac{5k}{4}=\frac{49k}{12}=49\)
\(\Rightarrow\frac{49k}{12}=49\)
\(\Rightarrow49k=49\cdot12\)
\(\Rightarrow k=12\)
Tự thay vào rồi tính x; y; z
b) \(\left(\frac{x}{2}\right)^3=\frac{x}{2}\cdot\frac{x}{2}\cdot\frac{x}{2}=\frac{x}{2}\cdot\frac{y}{3}\cdot\frac{z}{5}=\frac{xyz}{2\cdot3\cdot5}=\frac{810}{30}=27=3^3\)
\(\Rightarrow\frac{x}{2}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{5}=3\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=15\end{cases}}}\)
a)\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)
Với \(\frac{x}{\frac{3}{2}}=12\Rightarrow x=18\)
Với \(\frac{y}{\frac{4}{3}}=12\Rightarrow y=16\)
Với \(\frac{z}{\frac{5}{4}}=12\Rightarrow z=15\)
Vậy ...
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)
Có \(xyz=810\)
\(\Leftrightarrow2k.3k.5k=810\)
\(\Leftrightarrow k^3=27\)
\(\Leftrightarrow k=3\)
\(\Rightarrow\hept{\begin{cases}x=2k=2.3=6\\y=3k=3.3=9\\z=5k=5.3=15\end{cases}}\)
Vậy ...
