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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)
\(=\frac{15-4}{26}=\frac{11}{26}\)
c) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\)
\(17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)\)
\(17\frac{2}{31}=\frac{529}{31};6\frac{2}{31}=\frac{188}{31}\)
\(17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)\)
\(=\frac{529}{31}-\left(\frac{15}{17}+\frac{188}{31}\right)\)
\(=\frac{529}{31}-\frac{3661}{527}\)
\(=\frac{172}{17}\)
\(\left(\frac{29}{31}-\frac{7}{8}\right)-\left(\frac{28}{31}-4\right)\)
\(=\frac{15}{248}-\left(-\frac{96}{31}\right)\)
\(=\frac{783}{248}\)
\(\frac{1-\frac{6}{23}+\frac{6}{31}-\frac{6}{154}}{\frac{5}{2}-\frac{15}{23}+\frac{15}{31}-\frac{15}{154}}\)
\(=\frac{\frac{6}{6}-\frac{6}{23}+\frac{6}{31}-\frac{6}{154}}{\frac{15}{6}-\frac{15}{23}+\frac{15}{31}-\frac{15}{154}}\)
\(=\frac{6\left(\frac{1}{6}-\frac{1}{23}+\frac{1}{31}-\frac{1}{154}\right)}{15\left(\frac{1}{6}-\frac{1}{23}+\frac{1}{31}-\frac{1}{154}\right)}\)
\(=\frac{6}{15}=\frac{2}{5}\)
mk sửa đề tí nhé
thanks'