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\(\frac{25}{12}.\frac{23}{7}-\frac{25}{12}.\frac{12}{7}\)
\(=\frac{25}{12}.\left(\frac{23}{7}-\frac{12}{7}\right)\)\(\)
\(=\frac{25}{12}.\frac{11}{7}\)
\(=\frac{275}{84}\)
\(-\frac{6}{7}.\frac{7}{10}.\frac{11}{-6}.\left(-20\right)\)
\(=-\frac{3}{5}.\frac{-11}{6}.\left(-20\right)\)
\(=\frac{11}{10}.\left(-20\right)\)
\(=-22\)
Tính
\(\frac{12}{25}.\frac{23}{7}-\frac{12}{25}.\frac{12}{7}=\frac{12}{25}\left(\frac{23}{7}-\frac{12}{7}\right)\)
\(=\frac{12}{25}.\frac{11}{7}=\frac{132}{175}\)
\(-\frac{6}{11}.\frac{7}{10}.\frac{11}{-6}.\left(-20\right)\)
\(=\frac{-6.7.11.\left(-20\right)}{11.10.\left(-6\right)}=7.\left(-20\right)=-140\)
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
\(\dfrac{12}{25}\) . \(\dfrac{23}{7}\) - \(\dfrac{12}{7}\) . \(\dfrac{13}{25}\)
= \(\dfrac{12}{25}\). ( \(\dfrac{23}{7}\) - \(\dfrac{13}{7}\))
= \(\dfrac{12}{25}\) . \(\dfrac{10}{7}\)
= \(\dfrac{24}{35}\)