3x=2y

nen x/2=y/3

5x=2z

nên x/2=z/5

=>x/2=y/3=z/5

Đặt x/2=y/3=z/5=k

=>x=2k; y=3k; z=5k

Ta có: \(x^3+y^3-xyz=40\)

\(\Leftrightarrow8k^3+27k^3-30k^3=40\)

=>k=2

=>x=4; y=6; z=10

Ta có: \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21};5x+y-2z=28\)

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)

\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42};5x+y-2z\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

Suy ra: \(\frac{5x}{50}=2\Rightarrow x=2.50:5=10\)

\(\frac{y}{6}=2\Rightarrow y=2.6=12\)

\(\frac{2z}{42}=2\Rightarrow z=2.42:2=42\)

Vậy \(x=20;y=12;z=42\)

b) 3x = 2y; 7y = 5z; x - y + z =32

=> \(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\)

=> \(\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{x-y+z}{16}\)

=> \(\frac{x}{10}=2\Rightarrow x=20\)

=> \(\frac{y}{15}=2\Rightarrow y=30\)

=> \(\frac{z}{21}=2\Rightarrow z=42\)

d,

ta có:\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x.y.z}{2.3.5}=\frac{810}{30}=27\)

\(=>\frac{x}{2}=27=>x=2.27=54\)

\(=>\)\(\frac{y}{3}=277=>y=3.27=81\)

\(\frac{z}{5}=27=>z=5.27=135\)

vậy:x=54,y=81,z=135

a/

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)\(=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)\(\Rightarrow x=20;y=12;z=42\)

b/\(3x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{3};7y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+20}=2\)

\(\Rightarrow x=20;y=30;z=42\)

d) Đặt \(\frac{x}{2}=k\Rightarrow x=2k\); \(\frac{y}{3}=k\Rightarrow y=3k\); \(\frac{z}{5}=k\Rightarrow z=5k\)

Thay x=2k, y=3k, z=5k vào xyz=810 ta được:

\(2k.3k.5k=810\)

\(30k^3=810\)

\(k^3=\frac{810}{30}=27\)

\(\Rightarrow k=3\)

Do đó: x = 2k \(\Rightarrow\)x = 2.3=6

             y = 3k\(\Rightarrow\)y = 3.3=9

             z = 5k \(\Rightarrow\)z = 5.3=15

Vậy x=6; y=9; z=15

Từ đẳng thức : \(\frac{3x-2y}{5}=\frac{2z-5x}{3}=\frac{5y-3z}{2}\)

=> \(\frac{15x-10y}{5^2}=\frac{6z-15x}{3^2}=\frac{10y-6z}{2^2}\)

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

\(\frac{15x-10y}{5^2}=\frac{6z-15x}{3^2}=\frac{10y-6z}{2^2}=\frac{15x-10y+6z-15x+10y-6z}{5^2+3^2+2^2}=0\)

=> \(\hept{\begin{cases}15x=10y\\6z=15x\\10y=6z\end{cases}\Rightarrow\hept{\begin{cases}3x=2y\\2z=5x\\5y=3z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{z}{5}=\frac{x}{2}\\\frac{y}{3}=\frac{z}{5}\end{cases}\Rightarrow}\frac{x}{2}=\frac{y}{3}=\frac{z}{5}}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Khi đó : x² + 176 = yz 

<=> (2k)² - 15k² = -176

=> k²(4 - 15) = -176

=> k² = 16

=> k² = 4²

=> k = \(\pm\)4

Nếu k = 4 

=> \(\hept{\begin{cases}x=8\\y=12\\z=20\end{cases}}\)

Nếu k = - 4

=> \(\hept{\begin{cases}x=-8\\y=-12\\z=-20\end{cases}}\)