ho : B= x-y+z / x+2y -z va x/2 = y/5= z/7 va x+2y - z khac 0 

tìm x,y , z 

mày vớ vẩn

chẳng hỉu j

\(\frac{2}{x+y+z}=\frac{x}{2y+2z+1}=\frac{y}{2x+2z+1}=\frac{z}{2x+2y-2}=\frac{x+y+z}{4\left(x+y+z\right)}=\frac{1}{4}\)

\(\Rightarrow\hept{\begin{cases}2y+2z+1=4x\\2x+2z+1=4y\\x+y+z=8\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=\frac{17}{6}\\z=\frac{7}{3}\end{cases}}\)

Ta có: \(\left\{{}\begin{matrix}x\left(x+2y+3z\right)=-5\\y\left(x+2y+3z\right)=27\\z\left(x+2y+3z\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-5}=x+2y+3z\\\dfrac{y}{27}=x+2y+3z\\\dfrac{z}{5}=x+2y+3z\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{-5}=\dfrac{y}{27}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-27}{5}x\\z=-x\end{matrix}\right.\)

Ta có: \(x\left(x+2y+3z\right)=-5\Rightarrow x\left(x+2.\dfrac{-27}{5}x-3x\right)=-5\)

\(\Rightarrow\dfrac{-64}{5}x^2=-5\Rightarrow x^2=\dfrac{25}{64}\Rightarrow x=\dfrac{5}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{27}{5}x=-\dfrac{27}{8}\\z=-x=-\dfrac{5}{8}\end{matrix}\right.\)

vote cho mk nhé ok

a, Ta có : \(x:y:z=5:3:4\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{5}=\frac{y}{3}=\frac{z}{4}=\frac{x+2y-z}{5+6-4}=-\frac{126}{7}=-18\)

\(x=-90;y=-54;z=-72\)

b, \(5x=2y;3y=5z\Rightarrow\frac{x}{2}=\frac{y}{5};\frac{y}{5}=\frac{z}{3}\Rightarrow\frac{x}{2}=\frac{y}{5}=\frac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{3}=\frac{x+y+z}{2+5+3}=-\frac{970}{10}=-97\)

\(x=-194;y=-485;z=-291\)

c, \(3x=4y=5z\Rightarrow\frac{3x}{60}=\frac{4y}{60}=\frac{5z}{60}\Rightarrow\frac{x}{20}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau 

\(\frac{x}{20}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{20+15+12}=\frac{47}{47}=1\)

\(x=20;y=15;z=12\)