Đặt x/3=y/4=z/5=k

=>x=3k; y=4k; z=5k

Ta có: \(2x^2+y^2-z^2=9\)

\(\Leftrightarrow18k^2+16k^2-25k^2=9\)

\(\Leftrightarrow9k^2=9\)

\(\Leftrightarrow k^2=1\)

TH1: k=1

=>x=3; y=4; z=5

TH2: k=-1

=>x=-3; y=-4; z=-5

em chưa học đến :)

\(\dfrac{2x}{5}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

\(\Rightarrow\dfrac{2}{5}x=\dfrac{3}{4}y=\dfrac{4}{5}z\)

\(\Rightarrow\dfrac{2}{5}x.\dfrac{1}{12}=\dfrac{3}{4}y.\dfrac{1}{12}=\dfrac{4}{5}z.\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}\)

Đặt \(\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=30k\\y=16k\\z=15k\end{matrix}\right.\). Ta có:

\(x+y+z=49\)

\(\Rightarrow30k+16k+15k=49\)

\(\Rightarrow61k=49\)

\(\Rightarrow k=\dfrac{49}{61}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{49}{61}.30=\dfrac{1470}{61}\\y=\dfrac{49}{61}.16=\dfrac{784}{61}\\z=\dfrac{49}{61}.15=\dfrac{735}{61}\end{matrix}\right.\)

\(x^3=\dfrac{y^3}{8}=\dfrac{z^3}{27}\)

⇒\(x=\dfrac{y}{2}=\dfrac{z}{3}\)

⇒\(\dfrac{x^2}{1}=\dfrac{y^2}{4}=\dfrac{z^2}{9}=\dfrac{2x^2}{2}=\dfrac{7y^2}{28}=\dfrac{5z^2}{45}\)

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

\(\dfrac{x^2}{1}=\dfrac{y^2}{4}=\dfrac{z^2}{9}=\dfrac{2x^2}{2}=\dfrac{7y^2}{28}=\dfrac{5z^2}{45}=\dfrac{2x^2+7y^2+5z^2}{2+28-45}=\dfrac{-17}{-15}=\dfrac{17}{15}\)

⇒\(\dfrac{x^2}{1}=\dfrac{17}{15};\dfrac{y^2}{4}=\dfrac{17}{15};\dfrac{z^2}{9}=\dfrac{17}{15}\)

Còn lại bạn tự làm nha

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

<=>\(\dfrac{x}{2}=k\)=> x= 2k

<=>\(\dfrac{y}{3}\)\(=k\) => y= 3k

<=>\(\dfrac{z}{5}=k\) => z= 5k

Thay x= 2k, y=3k, z= 5k vào biểu thức xyz=810

Ta có: 2k . 3k . 5k = 810

<=> \(30k^3\) = 810

<=> \(k^3\) = 27

=> k = \(\sqrt[3]{27}\) = 3

\(\dfrac{x}{2}=3\) => x = 2 . 3 = 6

\(\dfrac{y}{3}=3\) => y = 3 . 3 = 9

\(\dfrac{z}{5}=3\) => z = 3 . 5 = 5

Vậy x = 6, y = 9, z = 15

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

\(=\dfrac{15x-10y}{25}=\dfrac{6z-15x}{9}=\dfrac{10y-6z}{4}\)

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

\(\dfrac{3x-2y}{5}=\dfrac{2z-5x}{3}=\dfrac{5y-3z}{2}=\dfrac{15x-10y}{25}=\dfrac{6z-15x}{9}=\dfrac{10y-6z}{4}\)

\(=\dfrac{15x-10y+6z-15x+10y-6z}{25+9+4}=0\)

⇒\(3x=2y\)⇒\(\dfrac{x}{2}=\dfrac{y}{3}\)

⇒\(2z=5x\)⇒\(\dfrac{x}{2}=\dfrac{z}{5}\)

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

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{9}=\dfrac{5z}{25}\)\(=\dfrac{2x+3y-5z}{6+9-25}=\dfrac{-60}{-10}=6\)

⇒\(\dfrac{x}{2}=6\)⇒\(x=12\)

⇒\(\dfrac{y}{3}=6\)⇒\(y=18\)

⇒\(\dfrac{z}{5}=6\)⇒\(z=30\)

Vậy \(x=12;y=18;z=30\)

\(\dfrac{2x}{5}=\dfrac{3y}{2}=\dfrac{5z}{7}\)

\(\Leftrightarrow28x=105y=50z\)

hay x/75=y/20=z/42

Đặt x/75=y/20=z/42=k

=>x=75k; y=20k; z=42k

Ta có: xyz=504000

\(\Leftrightarrow k^3\cdot63000=504000\)

\(\Leftrightarrow k=2\)

=>x=150; y=40; z=84