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a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a) \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\\ =\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\\ =\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{15}{26}-\dfrac{2}{13}\right)\\ =0+\left(\dfrac{15}{26}-\dfrac{4}{26}\right)\\ =0+\dfrac{11}{26}\\ =\dfrac{11}{26}\)
\(c)\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\\=\dfrac{-1}{23}\left ( \dfrac{66}{7}+\dfrac{88}{7}+1 \right )\\ =\dfrac{-1}{23}.23=-1\)
1) \(-\dfrac{5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =-\dfrac{5}{9}+\dfrac{14}{9}\cdot\dfrac{7}{20}\cdot\dfrac{1}{49}\\ =-\dfrac{5}{9}+\dfrac{1}{90}=\dfrac{-49}{90}\)
2) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right)\cdot\dfrac{24}{47}-\dfrac{51}{13}\cdot\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =-\dfrac{4}{13}\)
3) \(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{2}{5}\\ =1\)
Tìm x : 1) \(60\%x+0,4x+x:3=2\\ \Leftrightarrow\dfrac{3}{5}x+\dfrac{2}{5}x+\dfrac{1}{3}x=2\\ \Leftrightarrow\dfrac{4}{3}x=2\\ \Leftrightarrow x=\dfrac{3}{2}\)
Nốt nè bn
\(-2x-\dfrac{-3}{5}:\left(0,5\right)^2=-1\dfrac{1}{4}\\ \Leftrightarrow-2x+\dfrac{3}{5}:\dfrac{1}{4}=-\dfrac{5}{4}\\ \Leftrightarrow-2x+\dfrac{12}{5}=-\dfrac{5}{4}\\ \Leftrightarrow-2x=-\dfrac{73}{20}\\ x=-\dfrac{73}{40}\)
\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{2}{3}-\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{31}{60}\)
a) \(=\dfrac{-5}{12}.\dfrac{2}{7}+\dfrac{7}{12}.-\dfrac{3}{14}\)
\(=-\dfrac{5}{42}-\dfrac{1}{8}\)
= \(-\dfrac{41}{168}\)
b) \(=\dfrac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}\)
\(=\dfrac{1.1.3.\left(-2\right)}{\left(-1\right).2.1.\left(-2\right)}=\dfrac{3}{\left(-1\right).2}\)
\(=-\dfrac{3}{2}\)
a, \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
= \(\dfrac{28}{15}.\dfrac{25}{100}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
= \(\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{32-79}{60}\right).\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-47}{60}.\dfrac{24}{47}\)
= \(\dfrac{84}{60}+\dfrac{-24}{60}=\dfrac{60}{60}=1\)
b, \(\dfrac{\left(\dfrac{11^2}{200}+0,415\right):0,01}{\dfrac{1}{12}-37,25+3\dfrac{1}{6}}\)
= \(\dfrac{\left(\dfrac{121}{200}+\dfrac{415}{1000}\right):\dfrac{1}{100}}{\dfrac{1}{12}-\dfrac{3725}{100}+\dfrac{19}{6}}=\dfrac{\left(\dfrac{121}{200}+\dfrac{83}{200}\right).100}{\dfrac{1}{12}-\dfrac{149}{4}+\dfrac{19}{6}}\)
= \(\dfrac{\dfrac{51}{50}.100}{-34}=\dfrac{102}{-34}=-3\)
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
Sửa đề: \(\left(\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{13}{12}\right):\left(\dfrac{12}{23}+\dfrac{123}{234}+\dfrac{1234}{2345}\right)\)
\(=\left(\dfrac{4+9-13}{12}\right):\left(\dfrac{12}{23}+\dfrac{123}{234}+\dfrac{1234}{2345}\right)\)
=0