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a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)
`[-5]/6+8/3+[-29]/6 <= x <= [-1]/2+2+5/2`
`[-5]/6+16/6+[-29]/6 <= [6x]/6 <= [-3]/6+12/6+15/6`
`-5+16-29 <= 6x <= -3+12+15`
`-18 <= 6x <= 24`
`-18:6 <= 6x:6 <= 24:6`
`-3 <= x <= 4`
Vậy `-3 <= x <= 4`
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
a: \(=\dfrac{-12}{56}+\dfrac{35}{56}-\dfrac{28}{56}=-\dfrac{5}{56}\)
b: \(=\dfrac{5}{12}-\dfrac{4}{5}=\dfrac{25-48}{60}=\dfrac{-23}{60}\)
d: SỐ cần tìm là:
-24:3/8=-24x8:3=-64
a \(\dfrac{-5}{56}\)
b \(\dfrac{-23}{60}\)
c \(\dfrac{-23}{60}\)
d \(\dfrac{-1}{64}\)
2. \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-3}{2}\right)=\dfrac{-21}{4}.\dfrac{2}{7}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-15}{10}\right)=\dfrac{-3}{2}\)
\(\Leftrightarrow x.\dfrac{6}{5}=\dfrac{-3}{2}\)
\(\Leftrightarrow x=\dfrac{-3}{2}:\dfrac{6}{5}\)
\(\Leftrightarrow x=\dfrac{-3}{2}.\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{-5}{4}\)
3.\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=1\\2x-\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1+\dfrac{3}{4}\\2x=\left(-1\right)+\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=\dfrac{-7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}.\dfrac{1}{2}\\x=\dfrac{-7}{3}.\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{7}{6};\dfrac{-7}{6}\right\}\)
=>-3<n<=4
hay \(n\in\left\{-2;-1;0;1;2;3;4\right\}\)