Ta có: \(\dfrac{2}{3}x=\dfrac{1}{2}y=\dfrac{3}{5}z\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=y\cdot\dfrac{1}{2}=z\cdot\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{2}=\dfrac{z}{\dfrac{5}{3}}\)
mà x+y+z=-74
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{2}=\dfrac{z}{\dfrac{5}{3}}=\dfrac{x+y+z}{\dfrac{3}{2}+2+\dfrac{5}{3}}=\dfrac{-74}{\dfrac{31}{6}}=-\dfrac{444}{31}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{444}{31}\\\dfrac{1}{2}y=-\dfrac{444}{31}\\\dfrac{3}{5}\cdot z=-\dfrac{444}{31}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{444}{31}:\dfrac{2}{3}=\dfrac{-444}{31}\cdot\dfrac{3}{2}=\dfrac{-666}{31}\\y=-\dfrac{444}{31}:\dfrac{1}{2}=-\dfrac{444}{31}\cdot2=\dfrac{-888}{31}\\z=-\dfrac{444}{31}:\dfrac{3}{5}=-\dfrac{444}{31}\cdot\dfrac{5}{3}=-\dfrac{740}{31}\end{matrix}\right.\)
Vậy: (x,y,z)=\(\left(-\dfrac{666}{31};-\dfrac{888}{31};-\dfrac{740}{31}\right)\)
Ta có:\(\dfrac{2}{3}\)x=\(\dfrac{1}{2}\)y=\(\dfrac{3}{5}\)zhay x.\(\dfrac{2}{3}\)=y.\(\dfrac{1}{2}\)=z.\(\dfrac{3}{5}\)
=> \(\dfrac{x}{\dfrac{3}{2}}\)=\(\dfrac{y}{\dfrac{2}{1}}\)=\(\dfrac{z}{\dfrac{5}{3}}\)
Theo t/c mà giải
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