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a)
\(\dfrac{7}{24}+\dfrac{11}{6}\\ =\dfrac{7}{24}+\dfrac{11\times4}{6\times4}\\ =\dfrac{7}{24}+\dfrac{44}{24}\\ =\dfrac{7+44}{24}\\ =\dfrac{51}{24}\\=\dfrac{17}{8}\)
b)
\(3+\dfrac{5}{8}\\ =\dfrac{3\times8}{8}+\dfrac{5}{8}\\ =\dfrac{24}{8}+\dfrac{5}{8}\\ =\dfrac{29}{8}\)
c)
\(\dfrac{19}{28}-\dfrac{3}{7}\\ =\dfrac{19}{28}-\dfrac{3\times4}{7\times4}\\ =\dfrac{19}{28}-\dfrac{12}{28}\\ =\dfrac{7}{28}=\dfrac{1}{4}\)
d)
\(\dfrac{53}{21}-1\\ =\dfrac{53}{21}-\dfrac{21}{21}\\ =\dfrac{53-21}{21}\\ =\dfrac{32}{21}\)
a) \(\dfrac{7}{24}+\dfrac{11}{6}=\dfrac{7}{24}+\dfrac{44}{24}=\dfrac{7+44}{24}=\dfrac{51}{24}\)
b) \(3+\dfrac{5}{8}=\dfrac{24}{8}+\dfrac{5}{8}=\dfrac{24+5}{8}=\dfrac{29}{8}\)
c) \(\dfrac{19}{28}-\dfrac{3}{7}=\dfrac{19}{28}-\dfrac{12}{28}=\dfrac{19-12}{28}=\dfrac{7}{28}=\dfrac{1}{4}\)
d) \(\dfrac{53}{21}-1=\dfrac{53}{21}-\dfrac{21}{21}=\dfrac{53-21}{21}=\dfrac{32}{21}\)
\(a)\dfrac{5}{7}\times\dfrac{7}{15}\\ =\dfrac{5\times7}{7\times15}\\ =\dfrac{35}{105}=\dfrac{1}{3}\\ b)\dfrac{8}{21}\times3\\ =\dfrac{8\times3}{21}\\ =\dfrac{24}{21}=\dfrac{8}{7}\\ c)\dfrac{24}{17}:\dfrac{9}{17}\\ =\dfrac{24}{17}\times\dfrac{17}{9}\\ =\dfrac{24\times17}{17\times9}\\ =\dfrac{408}{153}=\dfrac{8}{3}\\ d)\dfrac{36}{5}:9\\ =\dfrac{36}{5}\times\dfrac{1}{9}\\ =\dfrac{36\times1}{5\times9}\\ =\dfrac{36}{45}=\dfrac{4}{5}\)
a) \(\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5\times7}{7\times15}=\dfrac{35}{105}=\dfrac{1}{3}\)
b) \(\dfrac{8}{21}\times3=\dfrac{8\times3}{21}=\dfrac{24}{21}=\dfrac{8}{7}\)
c) \(\dfrac{24}{17}:\dfrac{9}{17}=\dfrac{24}{17}\times\dfrac{17}{9}=\dfrac{24\times17}{17\times9}=\dfrac{24}{9}=\dfrac{8}{3}\)
d) \(\dfrac{36}{5}:9=\dfrac{36}{5}\times\dfrac{1}{9}=\dfrac{36}{5\times9}=\dfrac{36}{45}=\dfrac{4}{5}\)
C)
`7/19xx5/8xx (-19)/7`
`=7/19xx (-19/7)xx5/8`
`= -1xx5/8`
`=-5/8`
D)
`3/7xx9/11+3/7xx2/11`
`=3/7xx(9/11+2/11)`
`=3/7xx11/11`
`=3/7xx1`
`=3/7`
a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)
c,
\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)
\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)
d,
\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)
Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5
a)P=\(\dfrac{-2}{7}+\dfrac{14}{29}+\dfrac{12}{33}+\dfrac{15}{29}+\dfrac{21}{33}+\dfrac{2}{7}\)
=\(\left(\dfrac{-2}{7}+\dfrac{2}{7}\right)+\left(\dfrac{14}{29}+\dfrac{15}{29}\right)+\left(\dfrac{12}{33}+\dfrac{21}{33}\right)\)
=0+1+1=2
b)\(\dfrac{2}{7}.\dfrac{5}{19}+\dfrac{2}{7}.\dfrac{14}{19}+\dfrac{21}{19}-\dfrac{2}{7}.\dfrac{1}{5}\)
=\(\dfrac{2}{7}.\left(\dfrac{5}{19}+\dfrac{14}{19}-\dfrac{1}{5}\right)+\dfrac{21}{19}\)
=\(\dfrac{2}{7}.\dfrac{4}{5}+\dfrac{21}{19}=\dfrac{887}{665}\)
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
98775 - 32 x 85
=98775 -2720
=96055
67500 - 24 x 236
= 67500 -5664
=61836
568 + 101598 : 287
= 568 +354
=922
6875 + 980 -180
=7855 -180
=7675
\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)
\(=\dfrac{7}{10}-\dfrac{1}{2}\)
= \(\dfrac{1}{5}\)
\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)
\(=\dfrac{8}{11}+\dfrac{2}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}x\dfrac{8}{5}\)
\(=\dfrac{8}{30}\)
\(=\dfrac{4}{15}\)
\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)
\(=\dfrac{5}{12}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}-\dfrac{2}{12}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
a) \(\dfrac{3}{5}+\dfrac{7}{25}=\dfrac{15}{25}+\dfrac{7}{25}=\dfrac{15+7}{25}=\dfrac{22}{25}\)
b) \(\dfrac{8}{11}-\dfrac{19}{33}=\dfrac{24}{33}-\dfrac{19}{33}=\dfrac{24-19}{33}=\dfrac{5}{33}\)
c) \(\dfrac{16}{21}\times\dfrac{3}{5}=\dfrac{16\times3}{21\times5}=\dfrac{48}{105}=\dfrac{16}{35}\)
d) \(\dfrac{14}{41}\div\dfrac{7}{9}=\dfrac{14}{41}\times\dfrac{9}{7}=\dfrac{14\times9}{41\times7}=\dfrac{126}{287}=\dfrac{18}{41}\)