Đặt x/2=y/3=z/5=k

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

2x^2+y^2-z^2=34

=>2*4k^2+9k^2-25k^2=34

=>-8k^2=34

=>Loại

\(\dfrac{x}{3}=\dfrac{y}{2};\dfrac{x}{4}=\dfrac{z}{5}\) và \(x+y-z=10\)

Ta có:

\(\dfrac{x}{3}=\dfrac{y}{2}\Leftrightarrow\dfrac{x}{12}=\dfrac{y}{8};\dfrac{x}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{12}=\dfrac{z}{15}\)

\(\Rightarrow\dfrac{y}{8}=\dfrac{x}{12}=\dfrac{z}{15}\) và \(x+y-z=10\)

AD tính chất DTS bằng nhau ta có:

\(\dfrac{y}{8}=\dfrac{x}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{12+8-15}=\dfrac{10}{5}=2\)

+) \(\dfrac{y}{8}=2\Rightarrow y=16\)

+) \(\dfrac{x}{12}=2\Rightarrow x=42\)

+) \(\dfrac{z}{15}=2\Rightarrow z=30\)

Vậy \(x=42;y=16;z=30\)

c,\(\dfrac{x}{2}=\dfrac{y}{5};\dfrac{y}{3}=\dfrac{z}{2}\) và \(2x+3y-4z=34\)

Ta có:

\(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{15};\dfrac{y}{3}=\dfrac{z}{2}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)

Ta lại có:

\(\dfrac{2x}{12}=\dfrac{3y}{45}=\dfrac{4z}{40}\) và \(2x+3y-4z=34\)

AD tính chất DTS bằng nhau ta có:

\(\dfrac{2x}{12}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{12+45-40}=\dfrac{34}{17}=2\)

+) \(\dfrac{2x}{12}=2\Rightarrow x=12\)

+) \(\dfrac{3y}{45}=2\Rightarrow y=30\)

+) \(\dfrac{4z}{40}=2\Rightarrow z=20\)

Vậy \(x=12;y=30;z=20\)

\(\)

kcj

\(3x=2y=z\Rightarrow\frac{z}{6}=\frac{x}{2}=\frac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{z}{6}=\frac{x}{2}=\frac{y}{3}=\frac{x+y+z}{6+2+3}=\frac{99}{11}=9\)

\(\Rightarrow\hept{\begin{cases}z=54\\x=18\\y=27\end{cases}}\)

\(\frac{2x}{1}=\frac{-3y}{-1}=\frac{4z}{-2}\)

áp dụng tính chất dãy tỉ số bằng nhau  ta có

\(\frac{2x}{1}=\frac{-3y}{-1}=\frac{4z}{-2}=\frac{2x-3y+4z}{1+-1-2}=\frac{48}{-2}=-24\)

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

\(\frac{x}{2}=\frac{y}{5};\frac{y}{3}=\frac{z}{2}\) và 2x + 3y - 4z = 34 

\(\frac{x}{2}=\frac{y}{5}=\frac{1}{3}.\frac{x}{2}=\frac{1}{3}.\frac{y}{5}=\frac{x}{6}=\frac{y}{15}\)

\(\frac{y}{3}=\frac{z}{2}=\frac{1}{5}.\frac{y}{3}=\frac{1}{5}.\frac{z}{2}=\frac{y}{15}=\frac{z}{10}\)

\(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\) và 2x + 3y -4z = 34 

Theo tính chất dãy tỉ số bằng nhau: 

\(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\Rightarrow\frac{2x+3y-4z}{12+45-40}=\frac{34}{17}=2\)

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

\(\frac{y}{15}=2\Rightarrow y=2.15=30\)

\(\frac{z}{10}=2\Rightarrow z=2.10=20\)

Vậy...

làm thế nào đấy

1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

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