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Lời giải:
a)
\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+....+\frac{-1}{90}\)
\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+...+\frac{-1}{9.10}\)
\(=\frac{4-5}{4.5}+\frac{5-6}{5.6}+\frac{6-7}{6.7}+....+\frac{9-10}{9.10}\)
\(=\frac{1}{5}-\frac{1}{4}+\frac{1}{6}-\frac{1}{5}+\frac{1}{7}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{9}\)
\(=\frac{1}{10}-\frac{1}{4}=-\frac{3}{20}\)
b)
\(2B=5+\frac{8}{11}+\frac{3}{11}+\frac{1}{15}+\frac{13}{15.2}\)
\(=5+\frac{11-3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{15-2}{15.2}\)
\(=5+1-\frac{3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{1}{2}-\frac{1}{15}\)
\(=5+1+\frac{1}{2}=\frac{13}{2}\Rightarrow B=\frac{13}{4}\)
Lời giải:
a, Đặt \(A=\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
\(\Rightarrow A=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+...+\dfrac{-1}{9.10}\)
\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{5}-\dfrac{1}{5}+...-\dfrac{1}{9}+\dfrac{1}{10}\)
\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{10}\)
\(\Rightarrow A=\dfrac{-3}{20}\)
\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{9.10}\\ =\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{10}\\=\dfrac{-1}{4}+\dfrac{1}{10}\\=\dfrac{-5}{20}+\dfrac{2}{20}\\=\dfrac{-3}{20}\)
a) \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\\ =\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\\ =\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{15}{26}-\dfrac{2}{13}\right)\\ =0+\left(\dfrac{15}{26}-\dfrac{4}{26}\right)\\ =0+\dfrac{11}{26}\\ =\dfrac{11}{26}\)
\(c)\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\\=\dfrac{-1}{23}\left ( \dfrac{66}{7}+\dfrac{88}{7}+1 \right )\\ =\dfrac{-1}{23}.23=-1\)
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) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
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, \(\dfrac{\left(\dfrac{11^2}{200}+0,415\right):0,01}{\dfrac{1}{12}-37,25+3\dfrac{1}{6}}\)
= \(\dfrac{\left(\dfrac{121}{200}+\dfrac{415}{1000}\right):\dfrac{1}{100}}{\dfrac{1}{12}-\dfrac{3725}{100}+\dfrac{19}{6}}=\dfrac{\left(\dfrac{121}{200}+\dfrac{83}{200}\right).100}{\dfrac{1}{12}-\dfrac{149}{4}+\dfrac{19}{6}}\)
= \(\dfrac{\dfrac{51}{50}.100}{-34}=\dfrac{102}{-34}=-3\)
b: \(C=\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\cdot\dfrac{3-2-1}{6}=0\)