Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\left[\dfrac{13}{15}+\dfrac{13}{35}+\dfrac{13}{63}+\dfrac{13}{99}\right]=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\right]=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\right]=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\dfrac{52}{33}=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=40\)
hay x=40:3/2=80/3
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[13\left(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}\right)\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[\dfrac{13}{2}\cdot\dfrac{8}{33}\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-45=-5\)
=>2/3x=40
=>3x=1/20
hay x=1/60
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...)
\(a,\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.3.\left[\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left(\dfrac{1}{5-1}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.\dfrac{1}{3}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{3}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
\(b,1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
\(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x.\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+x+1-\dfrac{x}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)
=>\(x+1=1993\)
\(x=1993-1\)
\(x=1992\)
\(C=70.\left(131313.\left(\dfrac{1}{565656}+\dfrac{1}{727272}+\dfrac{1}{909090}\right)\right)\)
\(C=70.\left(131313.\dfrac{1}{235690}\right)\)
\(C=70.\dfrac{39}{70}\)
\(C=39\)
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) \(0,2\cdot\dfrac{15}{36}-\left(\dfrac{2}{5}+\dfrac{2}{3}\right):1\dfrac{1}{5}\)
\(=\dfrac{1}{12}-\dfrac{16}{15}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{12}-\dfrac{8}{9}\)
\(=\dfrac{-29}{36}\)
b) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{8}{15}+0,25\right)\cdot\dfrac{24}{27}\)
\(=\dfrac{28}{15}\cdot0,75-\dfrac{47}{60}\cdot\dfrac{24}{27}\)
\(=\dfrac{7}{5}-\dfrac{94}{135}\)
\(=\dfrac{19}{27}\)
c) \(5:\left(4\dfrac{3}{4}-1\dfrac{25}{28}\right)-1\dfrac{3}{8}:\left(\dfrac{3}{8}+\dfrac{9}{20}\right)\)
\(=5\cdot\dfrac{7}{20}-\dfrac{11}{8}\cdot\dfrac{40}{33}\)
\(=\dfrac{7}{4}-\dfrac{5}{3}\)
\(=\dfrac{1}{12}\)
A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)
A=\(\dfrac{1}{1}-\dfrac{1}{39}\)
A=\(\dfrac{38}{39}\)
còn lại tự làm do mình có việc chút
\(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\cdot\cdot\cdot\left(1+\dfrac{7}{180}\right)=\dfrac{16}{9}\cdot\dfrac{27}{20}\cdot\cdot\cdot\dfrac{187}{180}=\dfrac{2.8}{1\cdot9}\cdot\dfrac{3\cdot9}{2\cdot10}\cdot\cdot\cdot\dfrac{11\cdot17}{10\cdot18}=\dfrac{\left(2\cdot3\cdot...\cdot11\right)\cdot\left(8\cdot9\cdot...\cdot17\right)}{\left(1\cdot2\cdot...\cdot10\right)\cdot\left(9\cdot10\cdot...\cdot18\right)}=\dfrac{11\cdot8}{1\cdot18}=\dfrac{88}{18}=\dfrac{44}{9}\)
a: \(\dfrac{22}{33}+\dfrac{-444}{555}=\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{10}{15}-\dfrac{12}{15}=-\dfrac{2}{15}\)
b: \(\dfrac{-123}{246}+\dfrac{-131313}{151515}=\dfrac{-1}{2}+\dfrac{-13}{15}\)
\(=\dfrac{-15+\left(-26\right)}{30}=\dfrac{-41}{30}\)
a) 22/33 + (-444/555)
= 2/3 - 4/5
= -2/15
b) -123/246 + (-131313/151515)
= -1/2 - 13/15
= -41/30