\(4x=2y\Rightarrow2x=y\Rightarrow\frac{x}{1}=\frac{y}{2};7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)

\(\frac{x}{5}=\frac{y}{10}=\frac{z}{14}=\frac{x-y+z}{5-10+14}=-\frac{46}{9}\)

\(x=5.\frac{-46}{9}=-\frac{230}{9}\)

\(y=10.\frac{-46}{9}=-\frac{460}{9}\)

\(z=14.\frac{-46}{9}=-\frac{644}{9}\)

Ta có: \(4x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{4}\)\(\Rightarrow\frac{x}{15}=\frac{y}{20}\)

\(7y=5z\)\(\Rightarrow\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\)\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\)

Ta có: \(yz-2x^2=110\)

\(\Rightarrow20k.28k-2.\left(15k\right)^2=110\)

\(\Rightarrow560k^2-2.225k^2=110\)

\(\Rightarrow560k^2-450k^2=110\)

\(\Rightarrow k^2\left(560-450\right)=110\)

\(\Rightarrow110k^2=110\)

\(\Rightarrow k^2=1\)

\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

+) Khi k = 1, ta có: \(\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\Rightarrow\hept{\begin{cases}x=15.1\\y=20.1\\z=28.1\end{cases}}\Rightarrow\hept{\begin{cases}x=15\\y=20\\z=28\end{cases}}\)

+) Khi k = -1, ta có: \(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\Rightarrow\hept{\begin{cases}x=15.\left(-1\right)\\y=20.\left(-1\right)\\z=28.\left(-1\right)\end{cases}}\Rightarrow\hept{\begin{cases}x=-15\\y=-20\\z=-28\end{cases}}\)

Vậy...

Ta có: \(4x=3y\rightarrow\frac{x}{3}=\frac{y}{4}\rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)

          \(7y=5z\rightarrow\frac{y}{5}=\frac{z}{7}\rightarrow\frac{y}{20}=\frac{z}{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\left(k\varepsilonℕ^∗\right)\)

=> x  = 15k; y = 20k; z = 28k

Có: \(yz-2x^2=110\)

\(\Rightarrow20k\cdot28k-2\cdot(15k)^2=110\)

\(\Rightarrow560\cdot k^2-2\cdot225\cdot k^2=110\)

\(\Rightarrow560\cdot k^2-450\cdot k^2=110\)

\(\Rightarrow\left(560-450\right)\cdot k^2=110\)

\(\Rightarrow110\cdot k^2=110\)            \(\Rightarrow k^2=1\)

\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

\(x=15k\rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

\(y=20k\rightarrow\orbr{\begin{cases}y=20\\y=-20\end{cases}}\)

\(z=28k\rightarrow\orbr{\begin{cases}z=28\\z=-28\end{cases}}\)

Vậy...........................

thiếu chỗ \(\hept{\begin{cases}x=15k\\y=20k\end{cases}}\):)) thêm vào là \(\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\)

\(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\)

\(7y=3z\Rightarrow\frac{y}{3}=\frac{z}{7}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{28}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{28}=\frac{x-y+z}{9-12+28}=\frac{-46}{25}\)

Vậy \(\hept{\begin{cases}x=-16,56\\y=--22,08\\z=-51,52\end{cases}}\)

4x = 3y => x/3 = y/4 (1)

5y = 3z => y/3 = z/5 (2)

từ (1), (2) => 9 x = 12 y = 20 z và 2x - 3y + z = 6

áp dụng tính chất của dãy tỉ số bằng nhau, có:

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{9\cdot2-3\cdot12+20}=\frac{6}{2}=3\) 

suy ra: \(\frac{x}{9}=3=>x=9\cdot3=27\)

\(\frac{y}{12}=3=>y=12\cdot3=36\)

\(\frac{z}{20}=3=>z=20\cdot3=60\)

Vậy.....

đơn giản

x = 27

y = 36

z = 60

\(\left(y+2\right)x+\left(y+2\right)=15\Leftrightarrow\left(y+2\right)\left(x+1\right)=15\)