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6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
\(\dfrac{3}{7}.\left(-\dfrac{2}{5}\right).2\dfrac{1}{3}.20.\dfrac{19}{72}\)
\(=\dfrac{3}{7}.\left(-\dfrac{2}{5}\right).\dfrac{7}{3}.20.\dfrac{19}{72}\)
\(=\left(-\dfrac{6}{35}\right).\dfrac{7}{3}.20.\dfrac{19}{72}\)
\(=\left(-\dfrac{2}{5}\right).20.\dfrac{19}{72}\)
\(=\left(-8\right).\dfrac{19}{72}\)
\(=-\dfrac{19}{9}\)
1. Tính hợp lí
a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)
\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)
\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)
\(=1+\dfrac{-7}{19}\)
\(=\dfrac{12}{19}\)
b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)
\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)
\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)
\(=\dfrac{3}{5}.-1\)
\(=\dfrac{-3}{5}\)
d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)
\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)
\(=\dfrac{7}{4}.1\)
\(=\dfrac{7}{4}\)
Chúc bạn học tốt
a) 0,7+−719−(−0,3)0,7+19−7−(−0,3)
=710+−719+310=107+19−7+103
=(710+310)+−719=(107+103)+19−7
=1+−719=1+19−7
=1219=1912
b) 53.(−2,5):5635.(−2,5):65
=53.−52.65=35.2−5.56
=(53.65).−52=(35.56).2−5
=2.−52=2.2−5
=−5=−5
c) 0,6.−517−35.12170,6.17−5−53.1712
=35.−517−35.1217=53.17−5−53.1712
=35.(−517−1217)=53.(17−5−1712)
=35.−1=53.−1
=−35=5−3
d) 74.52−74.3247.25−47.23
=74.(52−32)=47.(25−23)
=74.1=47.1
=74=47
mình giúp rùi đó nhớ tick mình nha
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
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) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
2)
\(2+\dfrac{5}{7}+\left(\dfrac{\dfrac{3}{19}+\dfrac{3}{23}-\dfrac{3}{28}}{\dfrac{5}{19}+\dfrac{5}{23}-\dfrac{5}{28}}\right)\cdot x=\dfrac{20}{7}\\ \left[\dfrac{3\cdot\left(\dfrac{1}{19}+\dfrac{1}{23}-\dfrac{1}{28}\right)}{5\cdot\left(\dfrac{1}{19}+\dfrac{1}{23}-\dfrac{1}{28}\right)}\right]\cdot x=\dfrac{20}{7}-\dfrac{5}{7}-2\\ \dfrac{3}{5}x=\dfrac{15}{7}-2\\ \dfrac{3}{5}x=\dfrac{1}{7}\\ x=\dfrac{5}{21}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
\(=\dfrac{3}{7}\cdot\dfrac{7}{3}\cdot\dfrac{-2}{5}\cdot20\cdot\dfrac{19}{72}=-8\cdot\dfrac{19}{72}=-\dfrac{19}{9}\)