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\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\left[\dfrac{13}{15}+\dfrac{13}{35}+\dfrac{13}{63}+\dfrac{13}{99}\right]=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\right]=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{1}{3}-\dfrac{1}{11}\right)\right]=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}-\dfrac{780}{11}:\dfrac{52}{33}=-5\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=40\)
hay x=40:3/2=80/3
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) \(5:\left(4\dfrac{3}{4}-1\dfrac{25}{28}\right)-1\dfrac{3}{8}:\left(\dfrac{3}{8}+\dfrac{9}{20}\right)\)
\(=5\cdot\dfrac{7}{20}-\dfrac{11}{8}\cdot\dfrac{40}{33}\)
\(=\dfrac{7}{4}-\dfrac{5}{3}\)
\(=\dfrac{1}{12}\)
\(\dfrac{48}{25}\cdot\dfrac{27}{55}+2\dfrac{4}{9}\cdot\dfrac{14}{33}\)
\(=\dfrac{1296}{1375}+\dfrac{22}{9}\cdot\dfrac{14}{33}\\ =\dfrac{1296}{1375}+\dfrac{28}{27}\\ =\dfrac{34992}{37125}+\dfrac{38500}{37125}\\ =\dfrac{73492}{37125}\)
\(1\dfrac{19}{22}\cdot\left(\dfrac{47}{77}-\dfrac{16}{15}\right)\\ =\dfrac{41}{22}\cdot\dfrac{-527}{1155}\\ =\dfrac{-21607}{25410}\)
\(\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)-\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{307}{99}+\dfrac{37}{99}-\dfrac{1503}{299}\right)-0\\ =\dfrac{344}{99}-\dfrac{1053}{299}\\ =-\dfrac{107}{2277}\)
\(a,\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.3.\left[\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left(\dfrac{1}{5-1}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.\dfrac{1}{3}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{3}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
\(b,1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
\(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x.\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+x+1-\dfrac{x}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)
=>\(x+1=1993\)
\(x=1993-1\)
\(x=1992\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[13\left(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}\right)\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[\dfrac{13}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-\dfrac{780}{11}:\left[\dfrac{13}{2}\cdot\dfrac{8}{33}\right]=-5\)
\(\Leftrightarrow\dfrac{2}{3x}-45=-5\)
=>2/3x=40
=>3x=1/20
hay x=1/60
A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)
A=\(\dfrac{1}{1}-\dfrac{1}{39}\)
A=\(\dfrac{38}{39}\)
còn lại tự làm do mình có việc chút
a) Ta có:
\(\frac{232323}{999999}=\frac{23.10101}{99.10101}=\frac{23}{99}\)
Vì : \(\frac{23}{99}=\frac{23}{99}\\ =>\frac{23}{99}=\frac{232323}{999999}\)
b) \(-\frac{63}{84}=-\frac{3}{4}\\ \frac{65}{-91}=-\frac{5}{7}\)
Vì: \(-\frac{3}{4}< -\frac{5}{7}\\ =>-\frac{63}{84}< \frac{65}{-91}\)
c) Ta có: \(\frac{111}{115}=1-\frac{4}{115}\\ \frac{555}{559}=1-\frac{4}{559}\)
Vì: \(\frac{4}{115}>\frac{4}{559}\\ =>1-\frac{4}{115}< 1-\frac{4}{559}\\ =>\frac{111}{115}< \frac{555}{559}\)
a) \(\dfrac{23}{99}\) và \(\dfrac{232323}{999999}\)
* Giữ nguyên \(\dfrac{23}{99}\).
* Rút gọn \(\dfrac{232323}{999999}=\dfrac{23}{99}\).
Vì \(\dfrac{23}{99}=\dfrac{23}{99}\) nên \(\dfrac{23}{99}=\dfrac{232323}{999999}\)
Vậy \(\dfrac{23}{99}=\dfrac{232323}{999999}\).
b) \(\dfrac{-63}{84}\) và \(\dfrac{65}{-91}\)
\(\circledast\) Rút gọn:
\(\dfrac{-63}{84}=\dfrac{-3}{4}\) ; \(\dfrac{65}{-91}=\dfrac{-5}{7}\)
\(\circledast\) Quy đồng:
Mẫu chung: 28
\(\dfrac{-3}{4}=\dfrac{-3.7}{4\cdot7}=\dfrac{-21}{28}\)
\(\dfrac{-5}{7}=\dfrac{-5\cdot4}{7\cdot4}=\dfrac{-20}{28}\)
Vì \(\dfrac{-21}{28}< \dfrac{-20}{28}\) nên \(\dfrac{-63}{84}< \dfrac{65}{-91}\).
Vậy \(\dfrac{-63}{84}< \dfrac{65}{-91}\).
c) \(\dfrac{111}{115}\) và \(\dfrac{555}{559}\)
\(\dfrac{111}{115}=1-\dfrac{4}{115}\) ; \(\dfrac{555}{559}=1-\dfrac{4}{559}\)
Vì \(\dfrac{4}{115}>\dfrac{4}{559}\)
\(\Rightarrow\) \(1-\dfrac{4}{115}< 1-\dfrac{4}{559}\)
\(\Rightarrow\) \(\dfrac{111}{115}< \dfrac{555}{559}\)
Vậy \(\dfrac{111}{115}< \dfrac{555}{559}\).
A = \(\dfrac{17}{15}.\dfrac{-31}{125}.\dfrac{1}{2}.\dfrac{10}{17}.\dfrac{-1}{8}\)
= \(\dfrac{17.\left(-31\right).1.10.\left(-1\right)}{15.125.2.17.8}\)
= \(\dfrac{17.\left[\left(-31\right).\left(-1\right)\right].1.2.5}{5.3.125.17.4.2}\)
= \(\dfrac{31.1}{3.125.4}\)
= \(\dfrac{31}{1500}\)
B = \(\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}.\dfrac{11}{4}\right).\dfrac{8}{33}\)
= \(\left[\dfrac{11}{4}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\right].\dfrac{8}{33}\)
= \(\left(\dfrac{11}{4}.\dfrac{-9}{9}\right).\dfrac{8}{33}\)
= \(\left[\dfrac{11}{4}.\left(-1\right)\right].\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)
= \(\dfrac{-11}{4}.\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)
= \(\dfrac{\left(-11\right).4.2}{4.\left(-11\right)\left(-3\right)}\)
= \(\dfrac{2}{-3}\)
Ok nhá!
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