mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)

\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)

\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)

\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)

\(=\dfrac{100}{\dfrac{-25}{11}}\)

\(=-44\)

thiếu dấu kìa bạn

ở chỗ 10 và 5/6 ấy

tính giúp mình với

   \(\dfrac{(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)

= \(\dfrac{\left(13-2-10+\left(\dfrac{1}{4}-\dfrac{5}{27}-\dfrac{5}{6}\right)\right).\dfrac{46}{25}+\dfrac{67}{4}}{\dfrac{100}{21}:\left(-\dfrac{41}{21}\right)}\)

= \(\dfrac{\left(1-\dfrac{83}{108}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)

 = \(\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)

=  \(\dfrac{\dfrac{213}{4}+\dfrac{187}{4}}{-\dfrac{100}{4}}\)

= \(\dfrac{100}{-\dfrac{100}{4}}\)

= - 4 

a)\(-160\)

b)\(-7000\)

c)\(-\dfrac{1}{12}\)

d)\(-\dfrac{3}{2}\)

a: \(=\dfrac{-5}{8}-2\cdot\dfrac{5}{2}+\dfrac{1}{2}=-\dfrac{5}{8}-5+\dfrac{1}{2}=-\dfrac{41}{8}\)

a) 9/18 - (-7/12) + 13/32

= 13/12 + 13/32

= 143/96

b) (5/-8) + 14/25 - 6/10

= (-13/200) - 6/10

= -133/200

Chúc bạn học tốt!! ^^

a: \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{48}{96}+\dfrac{56}{96}+\dfrac{39}{96}\)

\(=\dfrac{143}{96}\)

b: \(\dfrac{-5}{8}+\dfrac{14}{25}-\dfrac{6}{10}\)

\(=\dfrac{-125}{200}+\dfrac{112}{200}-\dfrac{120}{200}\)

\(=\dfrac{-133}{200}\)

f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)

a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)

b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)

c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))