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S
24 tháng 7 2018
Giải:
a) \(\frac{1}{5}-\frac{2}{3}+2x=\frac{1}{2}\)
\(\Leftrightarrow2x=\frac{1}{2}-\left(\frac{1}{5}-\frac{2}{3}\right)\)
\(\Leftrightarrow2x=\frac{1}{2}-\frac{-7}{15}\)
\(\Leftrightarrow2x=\frac{11}{15}\)
\(\Leftrightarrow x=\frac{11}{15}:2\)
\(\Leftrightarrow x=\frac{11}{30}\)
b) \(4\left(\frac{1}{3}-3\right)+\frac{1}{2}=\frac{5}{6}+x\)
\(\Leftrightarrow\frac{-61}{6}=\frac{5}{6}+x\)
\(\Leftrightarrow x=\frac{-61}{6}-\frac{5}{6}\)
\(\Leftrightarrow x=\frac{-66}{6}=-11\)
MM
22 tháng 8 2019
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
NT
0
\(2\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow3-2x-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow2x-\left(-3+\frac{1}{3}\right)=7x-\frac{1}{4}\)
\(\Leftrightarrow2x-\left(-\frac{9}{3}+\frac{1}{3}\right)=7x-\frac{1}{4}\)
\(\Leftrightarrow2x-\left(-\frac{8}{3}\right)=7x-\frac{1}{4}\)
\(\Leftrightarrow2x+\frac{8}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow2x-7x=-\frac{8}{3}-\frac{1}{4}\)
\(\Leftrightarrow-5x=-\frac{32}{12}-\frac{4}{12}\)
\(\Leftrightarrow-5x=-3\)
\(\Leftrightarrow x=\left(-3\right):\left(-5\right)\)
\(\Leftrightarrow x=\frac{3}{5}\)
\(2\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
\(3-2x-\frac{1}{3}\)\(=7x-\frac{1}{4}\)
\(-2x-7x=-\frac{1}{4}-3+\frac{1}{3}\)
\(-9x=-\frac{35}{12}\)
\(x=-\frac{35}{12}:\left(-9\right)\)
\(x=\frac{35}{108}\)
Vậy \(x=\frac{35}{108}\)