Question

Tìm số hữu tỉ \(\dfrac{x+y}{y+z}\)\(.Biết\)\(\dfrac{y}{x}=2\) và \(\dfrac{z}{y}=3\)

Answer 1

\(\left\{{}\begin{matrix}\dfrac{y}{x}=2\Rightarrow y=2x\\\dfrac{z}{y}=3\Rightarrow z=3y\end{matrix}\right.\)

\(\Rightarrow\dfrac{x+y}{y+z}=\dfrac{x+2x}{y+3y}=\dfrac{3x}{4y}=\dfrac{3x}{8x}=\dfrac{3}{8}\)

Answer 2

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

\(\dfrac{y}{x}=2\Rightarrow y=2x\)

\(\dfrac{z}{y}=3\Rightarrow z=3y\)

\(\dfrac{x+y}{y+z}=\dfrac{x+2x}{2x+3y}=\dfrac{3x}{2x+6x}=\dfrac{3x}{8x}=\dfrac{3}{8}\)

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