\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)và \(xyz=810\)(1)

đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Rightarrow x=2k;y=3k;z=5k\)(2)

thay (2) vào (1), ta được:

\(xyz=2k\cdot3k\cdot5k=810\)

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\Leftrightarrow k=3\)

từ đó

\(\Rightarrow\hept{\begin{cases}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot5=15\end{cases}}\)

vậy x=6; y=9; z=15

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Rightarrow\hept{\begin{cases}x=\frac{2y}{3}\\z=\frac{5y}{3}\end{cases}}\)thế vào \(xyz=810\)ta đc: \(\frac{2y.5y.y}{3.3}=810\Leftrightarrow y^3=729\Leftrightarrow y=9\Rightarrow x=6;z=15\)

a, \(\frac{x}{3}=\frac{y}{4};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\Rightarrow x=27;y=36;z=60\)

b, \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

\(\Rightarrow x=18;y=24;z=30\)

c, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}=\frac{2x+3y-z-2-6+4}{4+9-4}=\frac{46}{9}\)

\(\Rightarrow x=\frac{101}{9};y=\frac{52}{3};z=\frac{220}{9}\)

d, Đặt \(x=2k;y=3k;z=5k\Rightarrow xyz=810\Rightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\Leftrightarrow k=3\)Với k = 3 thì \(x=6;y=9;z=15\)

Ta có:

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

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

=> \(\frac{x^3}{8}=\frac{y^3}{27}=\frac{z^3}{125}=\frac{810}{30}=27\)

=> \(\hept{\begin{cases}x^3=27.8=6^3\\y^3=27.27=9^3\\z^3=27.125=15^3\end{cases}}\)=> \(\hept{\begin{cases}x=6\\y=9\\z=15\end{cases}}\)

Vậy ...

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

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

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

Ta có

\(xyz=2k\cdot3k\cdot5k=810\)

\(\Rightarrow30k^3=810\)

\(\Rightarrow k^3=810:30=27\)

\(\Rightarrow k=3\)

Với \(k=3\)ta có

\(\hept{\begin{cases}x=2\cdot3\\y=3\cdot3\\z=5\cdot3\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=15\end{cases}}}\)

Vậy..................

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)và \(xyz=810\)

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

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

Thay \(\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)và \(xyz=810\)

Ta có : \(2k.3k.5k=810\)

            \(\left(2.3.5\right).\left(k.k.k\right)=810\)

           \(30.k^3=810\)

          \(k^3=810:30\)

         \(k^3=27\)

       \(k=3\)

Vì \(k=3\)

Ta có : \(\hept{\begin{cases}x=2.3=6\\y=3.3=9\\z=5.3=15\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=15\end{cases}}\)

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

=> xyz=2k.3k.5k=810

=> 30k³=810 =>k³=27 =>k=3

=>\(\hept{\begin{cases}x=2.3=6\\y=3.3=9\\z=5.3=15\end{cases}}\)

K còn cách khá ạ

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\\ \frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)

Từ (1);(2) Suy ra \(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất dãy tĩ số bằng nhau:

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}=\frac{2x}{18}=\frac{3y}{36}=\frac{z}{15}=\frac{2x-3y+z}{18-36+15}=\frac{6}{-3}=-2\)

Suy ra

x = (-2) . 9 = -18

y = (-2) . 12 = -24

z = (-2) . 15 = -30

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

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

Suy ra 

x = 2 . 10 = 20

y = 2 . 6 = 12

z = 2 . 21 = 42

Bài 1: Tìm x,y,z

a) Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

\(\Leftrightarrow\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{z-3}{4}\)

\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)

mà 2x+3y-z=50

nên áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{50-5}{9}=\frac{45}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}2x-2=5\cdot4\\3y-6=5\cdot9\\z-3=5\cdot4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=20+2=22\\3y=45+6=51\\z=20+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

Vậy: (x,y,z)=(11;17;23)

b) Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow2k\cdot3k\cdot5k=810\)

\(\Leftrightarrow30\cdot k^3=810\)

\(\Leftrightarrow k^3=27\)

hay k=3

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=3\cdot3=9\\z=5\cdot3=15\end{matrix}\right.\)

Vậy: (x,y,z)=(6;9;15)