| 7/5x+2/3|=|4/3x-1/4|
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Ta có: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)
=>\(\left[{}\begin{matrix}\dfrac{7}{5}x+\dfrac{2}{3}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{2}{3}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\dfrac{7}{5}x-\dfrac{4}{3}x=-\dfrac{1}{4}-\dfrac{2}{3}\\\dfrac{7}{5}x+\dfrac{4}{3}x=\dfrac{1}{4}-\dfrac{2}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\dfrac{1}{15}x=\dfrac{-11}{12}\\\dfrac{41}{15}x=\dfrac{-5}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{12}:\dfrac{1}{15}=-\dfrac{11}{12}\cdot15=-11\cdot\dfrac{5}{4}=-\dfrac{55}{4}\\x=-\dfrac{5}{12}:\dfrac{41}{15}=-\dfrac{5}{12}\cdot\dfrac{15}{41}=\dfrac{-5\cdot5}{4\cdot41}=\dfrac{-25}{164}\end{matrix}\right.\)