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\(\left(2017-\dfrac{5}{181}+\dfrac{1}{50}\right)-\left(4+\dfrac{3}{181}-\dfrac{2}{50}\right)+\left(1-\dfrac{8}{181}+\dfrac{3}{50}\right)\)
\(=2017-\dfrac{5}{181}+\dfrac{1}{50}-4-\dfrac{3}{181}+\dfrac{2}{50}+1-\dfrac{8}{181}+\dfrac{3}{50}\)
\(=2014+\dfrac{143}{4525}\)
\(=\dfrac{9113493}{4525}\)
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) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(50 số 1 )
= \(50+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)
A =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\)
⇒ 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\)
⇒ 2A - A =\(1-\dfrac{1}{2^{50}}\)
=50+1-\(\dfrac{1}{2^{50}}\)=51-\(\dfrac{1}{2^{50}}>3\)
Câu 1:
a) \(\left(\dfrac{4}{8}+\dfrac{1}{2}\right):\left(\dfrac{3}{5}-\dfrac{8}{7}\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{2}\right):\left(\dfrac{21}{35}-\dfrac{40}{35}\right)\)
\(=1:\dfrac{-19}{35}\)
\(=\dfrac{35}{-19}\)
b) \(-1\dfrac{3}{5}+0,34-50\%\)
\(=-\dfrac{8}{5}+\dfrac{34}{100}-\dfrac{50}{100}\)
\(=-\dfrac{160}{100}+\dfrac{34}{100}-\dfrac{50}{100}\)
\(=-\dfrac{44}{25}\)
Câu 2:
A = 357 - 62 - (-62 - 643)
A = 357 - 62 + 62 + 643
A = (357 + 643) - (62 - 62)
A = 1000 - 0
A = 1000
Câu 3:
\(\left(\dfrac{x}{5}-1\right):\left(-5\right)=\dfrac{1}{10}\)
\(\dfrac{x}{5}-1=\dfrac{1}{10}.\left(-5\right)\)
\(\dfrac{x}{5}-1=\dfrac{-1}{2}\)
\(\dfrac{x}{5}=\dfrac{-1}{2}+1\)
\(\dfrac{x}{5}=\dfrac{1}{2}\Rightarrow x=\dfrac{1.5}{2}=2,5\)
\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
a, \(-1\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{2}+2\dfrac{1}{6}\\ =-\dfrac{5}{3}+\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{13}{6}\\ =\dfrac{-5.4+3.3-1.6+13.2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
b, \(\dfrac{11}{50}\left(-17\dfrac{1}{2}\right)-\dfrac{11}{50}.82\dfrac{1}{2}\\ =\dfrac{11}{50}.\left(-17\dfrac{1}{2}-82\dfrac{1}{2}\right)=\dfrac{11}{50}.\left(-100\right)=-22\)
a) \(-1\dfrac{2}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(2\dfrac{1}{6}\)
=\(-\dfrac{5}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(\dfrac{13}{6}\)\()\)
=\(-\) \(\dfrac{20}{12}\) + \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) + \(\dfrac{26}{12}\)
= \((\)\(\dfrac{-20}{12}\) + \(\dfrac{26}{12}\) \()\) + \((\) \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) \()\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)
= \(\dfrac{3}{4}\)
b)\(\dfrac{11}{50}\) \((\) \(-17\dfrac{1}{2}\) \()\) \(-\) \(\dfrac{11}{50}\) .\(82\dfrac{1}{2}\)
= \(\dfrac{11}{50}\) . \(-\dfrac{35}{2}\) \(-\) \(\dfrac{11}{50}\) . \(\dfrac{165}{2}\)
= \(\dfrac{11}{50}\). \((\) \(-\dfrac{35}{2}\) \(-\) \(\dfrac{165}{2}\) \()\)
=\(\dfrac{11}{50}\). \(-\)\(100\)
= \(-22\)
Chúc bạn học thật tốt nha ! 
\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)
Thank!