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TN
6 tháng 8 2018
\(1\frac{1}{2}+x=\frac{3}{2}-7\)
<=> \(\frac{3}{2}+x=\frac{-11}{2}\)
<=> \(x=-7\)
\(\frac{1}{4}+\frac{1}{3}:3x=-5\)
<=> \(\frac{1}{3}:3x=\frac{-21}{4}\)
<=> \(3x=\frac{-4}{63}\)
<=> \(x=\frac{4}{189}\)
\(\frac{4}{5}.x=\frac{8}{35}\)
<=> \(x=\frac{2}{7}\)
\(\frac{2}{3x}-\frac{1}{4}=\frac{7}{1}\)
<=> \(\frac{2}{3x}=\frac{29}{4}\)
=> \(8=87x\)
<=> \(x=\frac{8}{87}\)
\(\frac{3}{5x}+\frac{1}{2}=\frac{1}{7}\)
<=> \(\frac{3}{5x}=\frac{-5}{14}\)
<=> \(-25x=42\)
<=> \(x=\frac{-42}{25}\)
\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16.\frac{2}{3}\right)=0\)
<=> \(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{-32}{3}=0\)
<=> \(\frac{43}{8}+x-\frac{173}{24}=\frac{-32}{3}\)
<=> \(\frac{43}{8}+x=\frac{-83}{24}\)
<=> \(x=\frac{-53}{6}\)
học tốt
TV
1
YN
4 tháng 2 2023
\(-\dfrac{3}{5}\times x+\dfrac{1}{2}=\dfrac{4}{5}\)
\(-\dfrac{3}{5}\times x=\dfrac{4}{5}-\dfrac{1}{2}\)
\(-\dfrac{3}{5}\times x=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}:-\dfrac{3}{5}\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
\(\dfrac{1}{2}\times x+\dfrac{3}{8}=\dfrac{7}{16}\)
\(\dfrac{1}{2}\times x=\dfrac{7}{16}-\dfrac{3}{8}\)
\(\dfrac{1}{2}\times x=\dfrac{1}{16}\)
\(x=\dfrac{1}{8}\)
Vậy \(x=\dfrac{1}{8}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}x=-\dfrac{1}{2}\)
\(x\times\left(\dfrac{2}{3}-\dfrac{5}{6}\right)=-\dfrac{1}{2}\)
\(x\times\dfrac{-1}{6}=-\dfrac{1}{2}\)
\(x=3\)
Vậy \(x=3\)
\(\dfrac{2}{5}x+\dfrac{3}{10}x=-\dfrac{1}{5}\)
\(x\times\left(\dfrac{2}{5}+\dfrac{3}{10}\right)=-\dfrac{1}{5}\)
\(x\times\dfrac{7}{10}=-\dfrac{1}{5}\)
\(x=-\dfrac{2}{7}\)
Vậy \(x=-\dfrac{2}{7}\).
HY
1
29 tháng 1
a: \(\dfrac{6}{7}:\left(\dfrac{2}{5}\cdot\dfrac{6}{7}\right)\)
\(=\dfrac{6}{7}:\dfrac{12}{35}\)
\(=\dfrac{6}{7}\cdot\dfrac{35}{12}=\dfrac{6}{12}\cdot\dfrac{35}{7}=\dfrac{5}{2}\)
b: \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)
\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)
\(=1-\dfrac{8}{9}=\dfrac{1}{9}\)
c: \(\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\)
\(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\)
\(=\dfrac{48+7-42}{56}=\dfrac{13}{56}\)
d: \(\dfrac{-1}{6}+\dfrac{2}{3}\cdot\dfrac{-3}{4}+\dfrac{4}{5}\)
\(=-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5-15+24}{30}=\dfrac{4}{30}=\dfrac{2}{15}\)
18 tháng 4 2023
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
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