\(\dfrac{-7}{8}.\dfrac{14}{23}-\dfrac{7}{8}.\dfrac{9}{23}+1\dfrac{8}{7}\)
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a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)
\(=\dfrac{15}{26}-\dfrac{4}{26}\)
\(=\dfrac{11}{26}\)
b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)
\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)
\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)
\(=\dfrac{-967}{63}\)
c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)
\(=-1\)
d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)
=0
a) \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
\(=\dfrac{1135}{23}-\left(\left(5+14\right)+\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\right)\)
\(=\dfrac{1135}{23}-\left(19+\dfrac{417}{736}\right)\)
\(=\dfrac{1135}{23}-19\dfrac{417}{736}\)
\(=\dfrac{1135}{23}-\dfrac{14401}{736}\)
\(=\dfrac{953}{32}\)
b) \(-\dfrac{3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=-\dfrac{1}{7}\cdot\dfrac{5}{3}-\dfrac{4}{3}\cdot\dfrac{1}{7}+\dfrac{17}{7}\)
\(=-\dfrac{5}{21}-\dfrac{4}{21}+\dfrac{17}{7}\)
\(=2\)
c) \(0,7\cdot2\dfrac{2}{3}\cdot20\cdot0,375\cdot\dfrac{5}{28}\)
\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=2\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
d) \(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(=\left(9\cdot\dfrac{3}{8}+\dfrac{303030}{69264}\right)+\dfrac{403}{100}\)
\(=\left(\dfrac{27}{8}+\dfrac{35}{8}\right)+\dfrac{403}{100}\)
\(=\dfrac{31}{4}+\dfrac{403}{100}\)
\(=\dfrac{589}{50}\)
P/s: Đánh dấu phẩy, dấu chấm (dấu nhân) cần rõ ràng (vì dấu chấm người ta sẽ hiểu là dấu nhân thay vì hiểu là dấu phẩy)
a) \(49\dfrac{8}{23}\)- \(\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
= \(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)+5\dfrac{7}{32}\)
=35+\(5\dfrac{7}{32}\)
=\(\dfrac{1287}{32}\)
b)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).\dfrac{9}{9}\right]+2\dfrac{3}{7}\)
=\(\left[\left(\dfrac{-3}{7}\right).1\right]+2\dfrac{3}{7}\)
=\(\left(\dfrac{-3}{7}\right)+2\dfrac{3}{7}\)
=2
c) 0,7.\(2\dfrac{2}{3}\).20.0.375.\(\dfrac{5}{28}\)
=0 (Vì có một thừ số là 0 nên nguyên cả tích là 0)
d)\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
=17+4,03
=21,03
a. \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
= \(\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
= \(\left(-1\right)+1=0\)
b. \(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
= \(\dfrac{22}{5}.2\dfrac{4}{33}-\dfrac{26}{5}+\dfrac{7}{5}\)
= \(\dfrac{28}{3}-\dfrac{33}{5}=\dfrac{41}{15}\)
c\(\dfrac{-7}{9}:\dfrac{8}{15}+\dfrac{-7}{9}.\dfrac{7}{15}+5\dfrac{7}{9}\)
= \(\dfrac{-56}{135}+\dfrac{-49}{135}+5\dfrac{7}{9}\)
= \(\dfrac{-7}{9}+5\dfrac{7}{9}=5\)
d. \(\left(16\dfrac{3}{8}-19\dfrac{3}{4}\right)-\left(12\dfrac{3}{8}-17\dfrac{3}{4}\right)\)
= \(16\dfrac{3}{8}-19\dfrac{3}{4}-12\dfrac{3}{8}+17\dfrac{3}{4}\)
= \(\left(16\dfrac{3}{8}-12\dfrac{3}{8}\right)-\left(19\dfrac{3}{4}-17\dfrac{3}{4}\right)\)
=\(4+2=6\)
a)\(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
=\(\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)=-1+1=0\)
b)\(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
=\(\dfrac{22}{5}.\dfrac{70}{33}-\dfrac{13}{5}.2+\dfrac{7}{5}\)
=\(\dfrac{28}{3}-\dfrac{26}{5}+\dfrac{7}{5}=\dfrac{83}{15}\)
Các câu sau tương tự
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
Bài 2 :
a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)
b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
a: =27/45-20/45=7/45
b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)
c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)
a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)
b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)
c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)
d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)
a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)
b:
Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)
\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)
\(=71-43+1+1\)
=28+2=30
\(\dfrac{-7}{8}\cdot\dfrac{14}{23}-\dfrac{7}{8}\cdot\dfrac{9}{23}+1\dfrac{8}{7}\)
\(=\dfrac{-7}{8}\left(\dfrac{9}{23}+\dfrac{14}{23}\right)+1+\dfrac{8}{7}\)
\(=-\dfrac{7}{8}+1+\dfrac{8}{7}=\dfrac{1}{8}+\dfrac{8}{7}=\dfrac{71}{56}\)
Lời giải:
$\frac{-7}{8}.\frac{14}{23}+\frac{-7}{8}.\frac{9}{23}+1+\frac{8}{7}$
$=\frac{-7}{8}(\frac{14}{23}+\frac{9}{23})+1+\frac{8}{7}$
$=\frac{-7}{8}.\frac{23}{23}+1+\frac{8}{7}$
$=\frac{-7}{8}.1+1+\frac{8}{7}$
$=1-\frac{7}{8}+\frac{8}{7}=\frac{1}{8}+\frac{8}{7}=\frac{71}{56}$