Giải bài 98 trang 49 - SGK Toán lớp 7 Tập 1
Tìm y, biết:
a) \(-\dfrac{3}{5}.y = \dfrac{21}{10};\)
b) \(y : \dfrac{3}{8} = -1\dfrac{31}{33};\)
c) \(1\dfrac{2}{5}.y + \dfrac{3}{7} = -\dfrac{4}{5};\)
d) \(-\dfrac{11}{12} . y + 0,25 = \dfrac{5}{6}.\)
a) \(-\dfrac{3}{5}.y = \dfrac{21}{10}\)
\( \Leftrightarrow y = \dfrac{21}{10} : \left(-\dfrac{3}{5}\right)\)
\( \Leftrightarrow y = \dfrac{21}{10} . \left(-\dfrac{5}{3}\right)\)
\( \Leftrightarrow y = -\dfrac{7}{2} = -3\dfrac{1}{2}\)
Vậy \(y = -3\dfrac{1}{2}\)
b) \(y : \dfrac{3}{8} = -1\dfrac{31}{33}\)
\( \Leftrightarrow y : \dfrac{3}{8} = -\dfrac{64}{33}\)
\( \Leftrightarrow y = -\dfrac{64}{33} .\dfrac{3}{8} \)
\( \Leftrightarrow y = -\dfrac{8}{11} \)
Vậy \( y = -\dfrac{8}{11} \)
c) \(1\dfrac{2}{5}.y + \dfrac{3}{7} = -\dfrac{4}{5}\)
\(\Leftrightarrow \dfrac{7}{5}.y + \dfrac{3}{7} = -\dfrac{4}{5}\)
\(\Leftrightarrow \dfrac{7}{5}.y = -\dfrac{4}{5} - \dfrac{3}{7} \)
\(\Leftrightarrow \dfrac{7}{5}.y = \dfrac{-28 - 15}{35} \)
\(\Leftrightarrow \dfrac{7}{5}.y = \dfrac{-43}{35} \)
\(\Leftrightarrow y = \dfrac{-43}{35} : \dfrac{7}{5}\)
\(\Leftrightarrow y = \dfrac{-43}{35} .\dfrac{5}{7} = \dfrac{-43}{49} \)
d) \(-\dfrac{11}{12} . y + 0,25 = \dfrac{5}{6}\)
\( \Leftrightarrow -\dfrac{11}{12} . y = \dfrac{5}{6} - 0,25\)
\( \Leftrightarrow -\dfrac{11}{12} . y = \dfrac{10}{12} - \dfrac{3}{12}\)
\( \Leftrightarrow -\dfrac{11}{12} . y = \dfrac{7}{12} \)
\( \Leftrightarrow y = \dfrac{7}{12} : \left(-\dfrac{11}{12}\right)\)
\( \Leftrightarrow y = \dfrac{7}{12} . \left(-\dfrac{12}{11}\right) = - \dfrac{7}{11} \)
Vậy \(y = -\dfrac{7}{11} \)
Lưu ý: \(\dfrac{a}{b} : \dfrac{c}{d} = \dfrac{a}{b} . \dfrac{d}{c}\)