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

Ta có: \(\left(13\frac14-2\frac{5}{27}-10\frac56\right)\cdot230\frac{1}{25}+46\frac34\)

\(=\left(13+\frac14-2-\frac{5}{27}-10-\frac56\right)\cdot230,04+46,75\)

\(=\left(1+\frac14-\frac{5}{27}-\frac56\right)\cdot230,04+46.75\)

\(=\left(\frac{108}{108}+\frac{27}{108}-\frac{20}{108}-\frac{90}{108}\right)\cdot230,04+46,75\)

\(=\frac{25}{108}\cdot230,04+46,75=53,25+46,75=100\)

Ta có: \(\left(1\frac37+\frac{10}{3}\right):\left(12\frac13-14\frac27\right)\)

\(=\left(\frac{10}{7}+\frac{10}{3}\right):\left(12+\frac13-14-\frac27\right)\)

\(=\left(\frac{30}{21}+\frac{70}{21}\right):\left(-2+\frac13-\frac27\right)\)

\(=\frac{100}{21}:\left(-2+\frac{7}{21}-\frac{6}{21}\right)=\frac{100}{21}:\left(-2+\frac{1}{21}\right)=\frac{100}{21}:\frac{-41}{21}=\frac{-100}{41}\)

Ta có: \(\frac{\left(13\frac14-2\frac{5}{27}-10\frac56\right)\cdot230\frac{1}{25}+46\frac34}{\left(1\frac37+\frac{10}{3}\right):\left(12\frac13-14\frac27\right)}\)

\(=100:\frac{-100}{41}=-41\)

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

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

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