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a) \(\frac{-28}{19}\). \(\frac{\left(-38\right)}{14}\)= 4
b) \(\frac{-21}{16}\). \(\frac{\left(-24\right)}{7}\)= \(\frac{9}{2}\)
c) \(\frac{12}{17}\). \(\frac{\left(-34\right)}{9}\)= \(\frac{-8}{3}\)
d) \(\frac{\left(-15\right)}{4}\): \(\frac{21}{10}\)= \(\frac{-25}{14}\)
e) \(-2\frac{1}{7}\): \(-1\frac{1}{14}\)= 2
nhớ ủng hộ mik nha mn
a.=4
b.=119/24
c.= -8/3
d.= -25/14
Mình chỉ làm đc từng này thôi
a) \(\frac{13}{26}-\frac{1}{3}-\frac{1}{2}+\frac{7}{21}\)
\(=\frac{1}{2}-\frac{1}{3}-\frac{1}{2}+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)
\(=0+0\)
\(=0\)
b) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\right)\)
\(=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{7}{17}+\frac{5}{17}\right)+\frac{6}{11}\)
\(=0+\frac{12}{17}+\frac{6}{11}\)
\(=\frac{132}{187}+\frac{102}{187}\)
\(=\frac{234}{187}\)
c) \(\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{12}{10}\right)\)
\(=\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{6}{5}\right)\)
\(=\frac{13}{5}+\frac{7}{16}-\frac{11}{16}+\frac{6}{5}\)
\(=\left(\frac{13}{5}+\frac{6}{5}\right)+\left(\frac{7}{16}-\frac{11}{16}\right)\)
\(=\frac{19}{5}+\left(\frac{-4}{16}\right)\)
\(=\frac{19}{5}-\frac{1}{4}\)
\(=\frac{76}{20}-\frac{5}{20}\)
\(=\frac{71}{20}\)
d) \(-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{21}{30}-\frac{5}{11}\right)\)
\(=-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{7}{10}-\frac{5}{11}\right)\)
\(=-\frac{3}{10}+\frac{6}{11}-\frac{7}{10}+\frac{5}{11}\)
\(=
\left(-\frac{3}{10}-\frac{7}{10}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)\)
\(=\frac{-10}{10}+\frac{11}{11}\)
\(=-1+1\)
\(=0\)
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
Ta có \(B=\left(5,25.12\frac{7}{10}+\frac{21}{4}.17\frac{3}{10}\right):\left(-6\frac{9}{16}\right)\)
\(=\left(\frac{21}{4}.12\frac{7}{10}+\frac{21}{4}.17\frac{3}{10}\right):\left(-\frac{105}{16}\right)\)
\(=\frac{21}{4}.\left(12\frac{7}{10}+17\frac{3}{10}\right):\left(-\frac{105}{16}\right)\)
\(=\frac{21}{4}.30:\frac{-105}{16}\)
\(=\frac{315}{2}.\left(\frac{-16}{105}\right)\)
\(=\frac{3.-8}{1}\)
\(=-24\)