g)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{32}{8}=4\)
\(+)\)\(\dfrac{x}{3}=4\Rightarrow x=4\times3=12\)
\(+)\)\(\dfrac{y}{5}=4\Rightarrow y=4\times5=20\)
Vậy \(x=12;y=20\)
h)
\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\left|-\dfrac{1}{5}\right|\)
\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\dfrac{1}{5}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{1}{2}-\dfrac{1}{5}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{10}-\dfrac{2}{10}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{10}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{3}{10}\) hoặc \(x-\dfrac{1}{3}=\dfrac{-3}{10}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{3}{10}\\x-\dfrac{1}{3}=\dfrac{-3}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}+\dfrac{1}{3}\\x=\dfrac{-3}{10}+\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{30}+\dfrac{10}{30}\\x=\dfrac{-9}{30}+\dfrac{10}{30}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{30}\\x=\dfrac{1}{30}\end{matrix}\right.\)
