Question

Tìm x,y,z

(x+y+1)/x = (x+z+2)/y = (x+y-3)/z = 1/(x+z+y) =6

và x+z+y=1/6

Answer 1

\(\frac{x+y+1}{x}=6\)

\(x+y+1=6x\)

\(y+6.\frac{1}{6}=5x\)

\(6x+7y+6z=5x\)

\(x+7y+6z=0\Rightarrow\frac{1}{6}+6y+5z=0\Rightarrow6y+5z=-\frac{1}{6}\)

\(\frac{x+z+2}{y}=6\Leftrightarrow13x+13z+6y=0\Leftrightarrow7x+7z=-1\Leftrightarrow x+z=-\frac{1}{7}\)

\(x+y+z-x-z=y=\frac{1}{6}-\left(-\frac{1}{7}\right)=\frac{13}{42}\)

\(6y+5z=-\frac{1}{6}\Leftrightarrow\frac{13}{7}+5z=-\frac{1}{6}\Leftrightarrow5z=-\frac{85}{42}\Leftrightarrow z=-\frac{17}{42}\)

\(x+y+z=\frac{1}{6}\Leftrightarrow x+\frac{13}{42}-\frac{17}{42}=\frac{1}{6}\Leftrightarrow x=\frac{1}{6}-\frac{13}{42}+\frac{17}{42}=\frac{11}{42}\)