Ta có :\(\frac{3}{x}+\frac{4}{y}+\frac{5}{z}=6\)

\(\Leftrightarrow\frac{6}{2x}+\frac{12}{3y}+\frac{20}{4z}=6\)

\(\Leftrightarrow\frac{6}{2x}+\frac{12}{2x}+\frac{20}{2x}=6\)

\(\Leftrightarrow\frac{6+12+20}{2x}=6\)

\(\Leftrightarrow\frac{19}{x}=6\)

\(\Leftrightarrow x=\frac{19}{6}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{2}{3}.\frac{19}{6}=\frac{19}{9}=y\)

\(\Leftrightarrow\frac{3}{4}y=\frac{3}{4}.\frac{19}{9}=\frac{19}{12}=z\)

Vậy \(\hept{\begin{cases}x=\frac{19}{6}\\y=\frac{19}{9}\\z=\frac{19}{12}\end{cases}}\)

\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}^{\left(1\right)}\)

\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}^{\left(2\right)}\)

\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{8}=\frac{z}{20}=\frac{y}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{8}=\frac{z}{20}=\frac{y}{15}=\frac{x+y+z}{8+20+15}=\frac{51}{43}\)

..... ( tới bước này bạn tự làm tiếp nhá )

\(\frac{z}{4}=\frac{y}{3}\Rightarrow\frac{z}{20}=\frac{y}{15}\)

\(\frac{x}{2}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{z}{20}\)

Suy ra:  \(\frac{x}{8}=\frac{y}{15}=\frac{z}{20}=\frac{x+y+z}{8+15+20}=\frac{51}{43}\)

Vậy \(x=8.\frac{51}{43}=\frac{408}{43}\)

\(y=15.\frac{51}{43}=\frac{765}{43}\)

\(z=20.\frac{51}{43}=\frac{1020}{43}\)

\(\frac{x-5}{3}=\frac{y-4}{4}=\frac{z-3}{5}=\frac{x-5+y-4+z-3}{3+4+5}=\frac{36-12}{12}=\frac{24}{12}=2\)

\(\Rightarrow\hept{\begin{cases}x-5=6\\y-4=8\\z-3=10\end{cases}}\Rightarrow\hept{\begin{cases}x=11\\y=12\\z=13\end{cases}}\)

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

Suy ra

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

\(y+z+1=2x\)(2)

\(x+z+2=2y\)(3)

\(x+y-3=2z\)(4)

(2)-(1) ta có

\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\Leftrightarrow y+z=\frac{1}{2}-\frac{1}{2}=0\)

\(y=-z\)

\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)

\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)

\(z=-\frac{5}{6}\)

Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\) và x² - y² + 2x² = 108

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

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4\)

\(\Rightarrow\frac{x^2}{4}=4\Rightarrow x=4\)

\(\Rightarrow\frac{y^2}{9}=4\Rightarrow y=6\)

\(\Rightarrow\frac{2z^2}{32}=4\Rightarrow z=8\)

theo dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)<=> \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)<=> \(\frac{x^2-y^2+2z^2}{4+9+32}=\frac{108}{45}=\frac{12}{5}\)

=> x=245

y=36/5

z= 48/5

\(\frac{x}{2}-2=\frac{y}{3}-2=\frac{z}{4}-2\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{27}{9}=3\)
\(\Rightarrow x=6,y=9,z=12\)

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

\(\frac{x-4}{2}=\frac{y-6}{3}=\frac{z-8}{4}=\frac{x+y+z-18}{2+3+4}=1\)

Ta có:\(\frac{x-4}{2}=1\Rightarrow x=6\)

\(\frac{y-6}{3}=1\Rightarrow y=9\)

\(\frac{z-8}{4}=1\Rightarrow z=12\)

Áp dụng tính chất tỉ lệ thức, ta có:

\(\frac{x-4}{2}=\frac{y-6}{3}=\frac{z-8}{4}=\frac{x+y+z-18}{9}=\frac{27-18}{9}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x-4}{2}=1\Leftrightarrow x=6\\\frac{y-6}{3}=1\Leftrightarrow y=9\\\frac{z-8}{4}=1\Leftrightarrow z=12\end{cases}}\)