Question

x/2=y/5;y/3=z/2 và 2x+3y-4z=34

Answer 1

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

áp dụng tính chất dảy tỉ số bằng nhau

ta có : \(\dfrac{2x+3y-4z}{2.6+3.15-4.10}=\dfrac{34}{12+45-40}=\dfrac{34}{17}=2\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{6}=2\\\dfrac{y}{15}=2\\\dfrac{z}{10}=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=12\\y=30\\z=20\end{matrix}\right.\) vậy \(x=12;y=30;z=20\)

Answer 2

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

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\Rightarrow\dfrac{2x}{12}=\dfrac{3y}{45}=\dfrac{4z}{40}\)

Và 2x + 3y - 4z = 34

Áp dụng t/c của dãy tỉ số = nhau ta có:

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

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2\cdot12}{2}=12\\y=\dfrac{2\cdot45}{3}=30\\z=\dfrac{2\cdot40}{4}=20\end{matrix}\right.\)

Answer 3

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{15}\left(1\right)\)

\(\dfrac{y}{3}=\dfrac{z}{2}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{10}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)

Ta lại có: \(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\Rightarrow\dfrac{2x}{36}=\dfrac{3y}{45}=\dfrac{4z}{40}\) và \(2x+3y-4z=34\)

AD t/c DTS bằng nahu ta có:

\(\dfrac{2x}{36}=\dfrac{3x}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{36+45-40}=\dfrac{34}{41}\)

......

Answer 4

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 \(2x+3y-4z=34\)

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

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

\(\left[{}\begin{matrix}\dfrac{x}{6}=2\\\dfrac{y}{15}=2\\\dfrac{z}{10}=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\y=30\\z=20\end{matrix}\right.\)

Vậy \(x=12;y=30;z=20\)