Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(1\dfrac{13}{15}.\left(-5\right)^2.3+\left(\dfrac{8}{15}-\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.25.3+\dfrac{13}{60}.\dfrac{24}{47}\)
\(=140+\dfrac{26}{235}=140\dfrac{26}{235}\)
b) \(\dfrac{\left(\dfrac{11^2}{200}+0,414:0,01\right)}{\dfrac{1}{12}-37.25+3\dfrac{1}{6}}\)
\(=\dfrac{\left(\dfrac{121}{200}-41,4\right)}{\dfrac{1}{12}-92519+\dfrac{19}{6}}\)
\(=\dfrac{2\dfrac{191}{207}}{-9251575}\)
Ta có :
\(\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{198}{2}+\dfrac{199}{1}\)
\(=\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(\dfrac{198}{2}+1\right)\left(\dfrac{199}{1}+1\right)-199\)\(=\dfrac{200}{199}+\dfrac{200}{199}+...+\dfrac{200}{2}+200-199\)
\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)
\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\right)\)
\(=200.A\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{200}\)
S sẽ có 30 số hạng. Nhóm thành 3 nhóm, mỗi nhóm 101 số hạng.
S= (1/31+1/32+...+1/40) + (1/41 + 1/42 +...+1/50) + (1/51 +1/52+...+1/60)
S < (1/30 + 1/30 +...+ 1/30) + ( 1/40 +1/40+...+1/40) + (1/50 +1/50+...+1/50)
S < 1/30 + 1/40 +1/50 ; S < 47/60 < 48/60 = 4/5 (1)
S > (1/40 + 1/40 +...=1/40) + (1/50 + 1/50 +...+1/50) + (1/60 +1/60+...+1/60)
S < 10/40 + 10/50 +10/60 ; S > 37/60 > 36/60 = 3/5 (2)
Tư (1) và (2) => 3/5 < S < 4/5
NHỚ TICK CHO MINK NHA, CHÚC BẠN HỌC TỐT
S=(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+...+\(\dfrac{1}{40}\))+(\(\dfrac{1}{41}\)+\(\dfrac{1}{42}\)+...+\(\dfrac{1}{50}\))+(\(\dfrac{1}{51}\)+\(\dfrac{1}{52}\)+...+\(\dfrac{1}{60}\))
=>\(\dfrac{10}{40}\)+\(\dfrac{10}{50}\)+\(\dfrac{10}{60}\)< S < \(\dfrac{10}{30}\)+\(\dfrac{10}{40}\)+\(\dfrac{10}{50}\)
=>\(\dfrac{37}{60}\)< S <\(\dfrac{47}{60}\)
=>\(\dfrac{3}{5}\)=\(\dfrac{36}{60}\)<\(\dfrac{37}{60}\)< S < \(\dfrac{47}{60}\)<\(\dfrac{48}{60}\)=\(\dfrac{4}{5}\)
=> \(\dfrac{3}{5}\)< S <\(\dfrac{4}{5}\)
a: \(=\dfrac{6}{18}+\dfrac{3}{18}+\dfrac{2}{18}+\dfrac{1}{18}=\dfrac{15}{18}=\dfrac{5}{6}\)
b: \(=\dfrac{7}{4}-\dfrac{5}{4}-\dfrac{2}{4}=0\)
c: \(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=1/2-1/10=4/10=2/5
a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)
\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)
=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)
=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)
b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)
=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)
c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)
\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)
\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)
=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)
\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)
Chúc bạn học tốt 
a)
\(A=\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+...+\dfrac{2}{52.53}\\ A=2\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{53.54}\right)\\ A=2.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{52}-\dfrac{1}{53}\right)\\ A=2\left(\dfrac{1}{3}-\dfrac{1}{53}\right)\\ A=\dfrac{100}{3.53}=\dfrac{100}{159}\)
b)
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2652}\\ B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{51.52}\\ B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{51}-\dfrac{1}{52}\\ B=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{50}{104}=\dfrac{25}{52}\)
câu c tương tự câu a
\(2\dfrac{x}{7}=\dfrac{14+x}{7}=\dfrac{75}{35}=\dfrac{15}{7}\\ \Rightarrow\left(14+x\right).7=15.7\\ 14+x=15\\ x=15-14=1\)
a) \(\dfrac{5}{3}+\dfrac{3}{-4}+\dfrac{7}{6}\) \(\left(MC:12\right)\)
\(=\dfrac{20}{12}+\dfrac{-9}{12}+\dfrac{14}{12}\)
\(=\dfrac{20+\left(-9\right)+14}{12}\)
\(=\dfrac{25}{12}\)
b) \(\dfrac{-1}{5}+\dfrac{5}{3}+\dfrac{-3}{2}\) \(\left(MC:30\right)\)
\(=\dfrac{-6}{30}+\dfrac{50}{30}+\dfrac{-45}{30}\)
\(=\dfrac{\left(-6\right)+50+\left(-45\right)}{30}\)
\(=\dfrac{-1}{30}\)
c) \(\dfrac{2}{7}+\dfrac{-7}{5}+\dfrac{-2}{35}\) \(\left(MC:35\right)\)
\(=\dfrac{10}{35}+\dfrac{-49}{35}+\dfrac{-2}{35}\)
\(=\dfrac{10+\left(-49\right)+\left(-2\right)}{35}\)
\(=\dfrac{-41}{35}\)
d) \(3+\dfrac{-7}{2}+\dfrac{-1}{5}\) \(\left(MC:10\right)\)
\(=\dfrac{30}{10}+\dfrac{-35}{10}+\dfrac{-2}{10}\)
\(=\dfrac{30+\left(-35\right)+\left(-2\right)}{10}\)
\(=\dfrac{-7}{10}\)
a) \(\dfrac{5}{3}+\dfrac{3}{-4}+\dfrac{7}{6}\)
\(=\dfrac{5}{3}+\dfrac{-3}{4}+\dfrac{7}{6}\)
\(=\) \(\dfrac{20}{12}+\dfrac{-9}{12}+\dfrac{14}{12}\)
\(=\dfrac{11}{12}+\dfrac{14}{12}\)
\(=\dfrac{25}{12}\)
b) \(\dfrac{-1}{5}+\dfrac{5}{3}+\dfrac{-3}{2}\)
\(=\dfrac{-6}{30}+\dfrac{50}{30}+\dfrac{-45}{30}\)
\(=\dfrac{44}{30}+\dfrac{-45}{30}\)
\(=\dfrac{-1}{30}\)
c) \(\dfrac{2}{7}+\dfrac{-7}{5}+\dfrac{-2}{35}\)
\(=\dfrac{10}{35}+\dfrac{-49}{35}+\dfrac{-2}{35}\)
\(=\dfrac{-39}{35}+\dfrac{-2}{35}\)
\(=\dfrac{-41}{35}\)
d) \(3+\dfrac{-7}{2}+\dfrac{-1}{5}\)
\(=\dfrac{3}{1}+\dfrac{-7}{2}+\dfrac{-1}{5}\)
\(=\dfrac{30}{10}+\dfrac{-35}{10}+\dfrac{-2}{10}\)
\(=\dfrac{-5}{10}+\dfrac{-2}{10}\)
\(=\dfrac{-7}{10}\)
ctv olm có mặt ạ
A = \(\dfrac{3}{24}\) + \(\dfrac{3}{40}\) + \(\dfrac{3}{60}\) + \(\dfrac{3}{84}\) + \(\dfrac{3}{112}\)
A = 3. ( \(\dfrac{2}{2.24}\) + \(\dfrac{2}{2.40}\) + \(\dfrac{2}{2.60}\)+ \(\dfrac{2}{2.84}\) + \(\dfrac{2}{2.112}\))
A = 3 .( \(\dfrac{2}{48}\) + \(\dfrac{2}{80}\) + \(\dfrac{2}{120}\) + \(\dfrac{2}{168}\) + \(\dfrac{2}{224}\))
A = 3. ( \(\dfrac{2}{6.8}\) + \(\dfrac{2}{8.10}\) + \(\dfrac{2}{10.12}\) + \(\dfrac{2}{12.14}\)+ \(\dfrac{2}{14.16}\))
A = 3.( \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\)- \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{16}\))
A = 3.(\(\dfrac{1}{6}\) - \(\dfrac{1}{16}\))
A = 3. \(\dfrac{5}{48}\)
A = \(\dfrac{15}{48}\)
A = \(\dfrac{5}{16}\)