\(h,\left|x-2,3\right|+\left|3,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-2,3\right|\ge0\\\left|3,5-x\right|\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2,3=0\\3,5-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2,3\\x=3,5\end{matrix}\right.\)
Vậy không có giá trị của x thỏa mãn đề.
\(k,\Leftrightarrow\dfrac{3}{2}x-\dfrac{1}{3}x=-\dfrac{1}{4}+\dfrac{2}{5}\\ \Leftrightarrow\dfrac{7}{6}x=\dfrac{3}{20}\Leftrightarrow x=\dfrac{9}{70}\)
k. \(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{3}x-\dfrac{1}{4}\)
<=> \(\dfrac{3}{2}x-\dfrac{1}{3}x=\dfrac{2}{5}-\dfrac{1}{4}\)
<=> \(\dfrac{7}{6}x=\dfrac{3}{20}\)
<=> \(\dfrac{3}{\dfrac{\dfrac{20}{7}}{6}}=\dfrac{1}{128}\)
h) \(\left|x-2,3\right|+\left|x-3,5\right|=0\)
Do \(\left|x-2,3\right|,\left|3,5-x\right|\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2,3=0\\3,5-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2,3\\x=3,5\end{matrix}\right.\)(vô lý)
Vậy \(S=\varnothing\)
k) \(\Leftrightarrow\dfrac{7}{6}x=\dfrac{3}{20}\)
\(\Leftrightarrow x=\dfrac{9}{70}\)
k: ta có: \(\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{3}x-\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{6}=\dfrac{3}{20}\)
hay \(x=\dfrac{9}{70}\)
