Question

5x = 6y = 8z và 2x +y + z = -83                                                                                         giúp mik vs 

Answer 1

Đề thiếu rồi em ơi !

Answer 2

có \(5x=6y=>x=\dfrac{6y}{5}\)

có \(6y=8z=>z=\dfrac{6y}{8}\)

thế x,z vào \(2x+y+z=-83\)

\(=>\dfrac{12y}{5}+y+\dfrac{6y}{8}=-83=>y=-20\)

\(=>x=\dfrac{6\left(-20\right)}{5}=-24\)

\(=>z=\dfrac{6\left(-20\right)}{8}=-15\)

Answer 3

Answer 4

\(5x=6y=8z\)

nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{6}}=\dfrac{z}{\dfrac{1}{8}}\)

hay \(\dfrac{2x}{\dfrac{2}{5}}=\dfrac{y}{\dfrac{1}{6}}=\dfrac{z}{\dfrac{1}{8}}=\dfrac{2x+y+z}{\dfrac{2}{5}+\dfrac{1}{6}+\dfrac{1}{8}}=\dfrac{-83}{\dfrac{83}{120}}=-120\)

\(\Leftrightarrow x=-24;y=-20;z=-15\)