Tìm x,y,x biết
\(a.\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và \(x+y+z=49\)
\(b.\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và \(2x+3y-z=186\)
\(c.x:y:z=12:9:5\)và \(xyz=20\)
\(d.\frac{10}{x-5}=\frac{6}{y-9}=\frac{14}{z-21}\)và \(xyz=6720\)
\(e.\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\)và \(2x^3-1=15\)
\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và x + y + z = 49
Ta có : \(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{2}+\frac{5}{4}}=\frac{49}{\frac{19}{4}}=49\cdot\frac{4}{19}=\frac{196}{19}\)
Vậy : \(\hept{\begin{cases}\frac{x}{\frac{3}{2}}=\frac{196}{19}\\\frac{y}{\frac{4}{2}}=\frac{196}{19}\\\frac{z}{\frac{5}{4}}=\frac{169}{14}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{294}{19}\\y=\frac{392}{19}\\z=\frac{245}{19}\end{cases}}\)
\(b,\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và 2x + 3y - z = 186
Ta có : \(\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\Leftrightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
Vậy : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)
\(c,x:y:z=12:9:5\)và xyz = 20
Ta có : \(x:y:z=12:9:5\)hay \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)
Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Leftrightarrow\hept{\begin{cases}x=12k\\y=9k\\z=5k\end{cases}}\)
\(\Leftrightarrow xyz=12k\cdot9k\cdot5k=540k^3\)
\(\Leftrightarrow xyz=540k^3\)
\(\Leftrightarrow540k^3=xyz\)
\(\Leftrightarrow540k^3=20\)
\(\Leftrightarrow k^3=\frac{20}{540}=\frac{1}{27}\Leftrightarrow k=\frac{1}{3}\)
Vậy : \(\hept{\begin{cases}x=12\cdot\frac{1}{3}\\y=9\cdot\frac{1}{3}\\z=5\cdot\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=\frac{5}{3}\end{cases}}\)
