\(\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow z=\dfrac{5y}{6}\)

mà \(z-y=40\)

\(\Rightarrow\dfrac{5y}{6}-y=40\)

\(\Rightarrow-\dfrac{y}{6}=40\)

\(\Rightarrow y=-240\Rightarrow z=40+y=40-240=-200\)

\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow x=\dfrac{3y}{4}=\dfrac{3.\left(-240\right)}{4}=-180\)

Vậy \(\left\{{}\begin{matrix}x=-180\\y=-240\\z=-200\end{matrix}\right.\)

\(\hept{\begin{cases}x-y=-9\\y-z=-10\\z+x=11\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y-9\\z=10+y\\10+y+y-9=11\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=y-9\\z=10+y\\2y=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=-4\\z=15\\y=5\end{cases}}}\)

\(\Rightarrow\frac{-1}{2}=\frac{x}{-10}=-\frac{7}{y}=\frac{z}{-21}\)

\(\Rightarrow x=-\frac{1}{2}.-10=5\)

\(y=-\frac{7}{-\frac{1}{2}}=14\)

\(z=-\frac{1}{2}.21=-\frac{21}{2}\)

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\)(*)

\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}\)(Dãy tỉ số bằng nhau)

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

\(\Rightarrow\frac{1}{x+y+z}=2\Leftrightarrow x+y+z=\frac{1}{2}\)

Thay vào (*), ta có:

\(\frac{\left(\frac{1}{2}-x\right)+1}{x}=\frac{\left(\frac{1}{2}-y\right)+2}{y}=\frac{\left(\frac{1}{2}-z\right)-3}{z}=2\)

\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{2}-x\\2y=\frac{5}{2}-y\\2z=-\frac{5}{2}-z\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{5}{2}\\3z=-\frac{5}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}.\)

Đặt  thì a + b + c = x + y + z

X= -1

Y= -2

Z= -3

T= -4

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