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a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
`7 xx 3/14 -1/14`
`= 21/14 -1/14`
`= 20/14`
`=10/7`
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`3/2 + 7/4 xx 2/5`
`= 3/2 + 14/20`
`= 3/2 + 7/10`
`= 15/10 +7/10`
`= 23/10`
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`9/8 : 3 + 7/3 : 3`
`= 9/8 xx 1/3 + 7/3 xx 1/3`
`=1/3 xx ( 9/8 + 7/3)`
`= 1/3 xx 83/24`
`= 83/72`
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`2 : 3/4 - 5/8 xx 4/3`
`= 2 xx 4/3 - 5/8 xx 4/3`
`= 4/3 xx ( 2-5/8)`
`= 4/3 xx ( 16/8 -5/8)`
`= 4/3 xx 11/8`
`= 44/24`
`=11/6`
` @ \color{Red}{sushiteam}`
b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)
=>x-1/2=27
hay x=55/2
c: =>1/2x-3/4=42/63=2/3
=>1/2x=17/12
hay x=17/6
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
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: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
\(4+\dfrac{3}{5}=\dfrac{20}{5}+\dfrac{3}{5}=\dfrac{23}{5}\)
\(\dfrac{1}{5}+\dfrac{2}{7}=\dfrac{7}{35}+\dfrac{10}{35}=\dfrac{17}{35}\)
\(\dfrac{4}{7}\times\dfrac{14}{3}=\dfrac{56}{21}=\dfrac{8}{3}\)
\(\dfrac{5}{6}:\dfrac{4}{9}=\dfrac{5}{6}\times\dfrac{9}{4}=\dfrac{15}{8}\)
\(4+\dfrac{3}{5}=\dfrac{20}{5}+\dfrac{3}{5}=\dfrac{23}{5}\)
\(\dfrac{1}{5}+\dfrac{2}{7}=\dfrac{7}{35}+\dfrac{5}{35}=\dfrac{12}{35}\)
\(\dfrac{4}{7}\times\dfrac{14}{3}=\dfrac{56}{21}=\dfrac{8}{3}\)
\(\dfrac{5}{6}:\dfrac{4}{9}=\dfrac{5}{6}\times\dfrac{9}{4}=\dfrac{15}{8}\)