Question

[Mai Tùng Dương]

Tìm x,y,z biết

\(\dfrac{x}{y}\) =\(\dfrac{7}{20}\);\(\dfrac{y}{z}\)=\(\dfrac{5}{8}\) và 2x +5y -2z = 100

Answer 1

[Phạm Ngân Hà]

\(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\) (1)

\(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\) (2)

Từ (1) và (2) suy ra \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

\(\Rightarrow\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x+5y-2z}{14+100-64}=\dfrac{100}{50}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.20=40\\z=2.32=64\end{matrix}\right.\)

Vậy...

Answer 2

[Hebico may mắn]

Ta có : \(\dfrac{x}{y}\) = \(\dfrac{7}{20}\) \(\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\) ( 1)

Ta có : \(\dfrac{y}{z}=\dfrac{5}{8}\) \(\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{y}{5}.\dfrac{1}{4}=\dfrac{z}{8}.\dfrac{1}{4}\)

\(\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\) (2)

Từ (1) và (2)

\(\Rightarrow\) \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

Đặt \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=k\)

\(\Rightarrow x=7k\) ; \(y=20k\) ; \(z=32k\)

Thay \(x=7k\) ; \(y=20k\) ; \(z=32k\) vào \(2x+5y-2z=100\)

\(\Rightarrow2.\left(7k\right)+5.\left(20k\right)-2.\left(32k\right)\) \(=100\)

\(\Rightarrow\)\(14k+100k-64k=100\)

\(\Rightarrow k.\left(14+100-64\right)=100\)

\(\Rightarrow k.50=100\)

\(\Rightarrow k=100:50\) \(\Rightarrow k=2\)

\(\Rightarrow x=7k=7.2=14\)

\(\Rightarrow y=20k=20.2=40\)

\(\Rightarrow z=32k=32.2=64\)

Vậy \(x=14\) ; \(y=40\) ;\(z=64\)