\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và y - x + z = -196
Tìm x, y , z biết :
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và \(-x+z=-196\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và \(z-x=-196\)
Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)
\(\Leftrightarrow\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\)
\(\Leftrightarrow\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}\)
mà z-x=-196
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{18x}{33}=\frac{18y}{4}=\frac{18z}{5}=\frac{18z-18x}{5-33}=\frac{18\left(z-x\right)}{-28}=\frac{-18\cdot196}{-28}=126\)
Do đó:
\(\left\{{}\begin{matrix}\frac{6x}{11}=126\\\frac{9}{2}y=126\\\frac{18z}{5}=126\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=1386\\y=28\\18z=630\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=231\\y=28\\z=35\end{matrix}\right.\)
Vậy: (x,y,z)=(231;28;35)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}zvà-x+z=-196\)
tìm x,y,z biết
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}zvà-x+z=-196\)
b, 2x=3y=5x=> I x+y-zI=96
\(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{-x+z}{-\frac{11}{6}+\frac{5}{18}}=-\frac{196}{-\frac{14}{9}}=126\)
x=\(126.\frac{11}{6}=231\)
y=\(126.\frac{2}{9}=28\)
z=\(126.\frac{5}{18}=35\)
b) \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{96}{19}\)hoặc \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=-\frac{96}{19}\)
=> ...........
Tìm x,y,z
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và -x + y + z = -120
\(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{-6x}{-11}=\frac{9y}{2}=\frac{18z}{5}\Rightarrow\frac{-x}{\frac{-11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)
\(\Rightarrow\frac{-x+y+z}{\frac{-11}{6}+\frac{2}{9}+\frac{5}{18}}=\frac{-120}{\frac{-4}{3}}=90\)
\(-x=90\times\frac{-11}{6}=-165\Rightarrow x=165\)
\(y=90\times\frac{2}{9}=20\)
\(z=90\times\frac{5}{18}=25\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và x+y+z
\(\frac{6}{11}.x=\frac{9}{2}.y=\frac{18}{5}.z\)và x+y-z= -120. x+y+z=?
Tìm x,y,z biết \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và -x+y+z=-120
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
=>x=165,y=20,z=25
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
Ba số x,y,z thỏa mãn: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và y-x+z=-120. Tính x+y+z=...