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\(a,\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{10}:\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{12}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{5}{4}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{11}{6}\)
\(\Leftrightarrow x=\dfrac{7}{12}-\dfrac{11}{6}\)
\(\Leftrightarrow\dfrac{-5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=\frac{-1}{2}+\frac{3}{4}=\frac{-2}{4}+\frac{3}{4}=\frac{1}{4}\)
\(\frac{x}{5}=\frac{5}{6}-\frac{19}{30}=\frac{25}{30}-\frac{19}{30}=\frac{6}{30}=\frac{1}{5}\)
*\(1\div2\)không phải \(\frac{1}{2}\)à?*
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
a, \(\frac{31}{12}-(\frac{2}{5}+x)=\frac{2}{3}\)
\(\Rightarrow\frac{2}{5}+x=\frac{31}{12}-\frac{2}{3}=\frac{23}{12}\)
\(\Rightarrow\frac{23}{12}-\frac{31}{12}=\frac{-8}{12}=\frac{-2}{3}\)
Câu b để mk làm sau
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{32^5\cdot81^4}{18^6\cdot16^5}=\frac{4^5\cdot8^5\cdot\left(9^2\right)^4}{2^6\cdot9^6\cdot\left(4^2\right)^5}=\frac{\left(2^2\right)^5\cdot2^5\cdot3^5\cdot\left(3^{2^2}\right)^4}{2^6\cdot\left(3^2\right)^6\cdot\left(2^{2^2}\right)^5}=\frac{2^{10}\cdot2^5\cdot3^5\cdot3^{16}}{2^6\cdot3^{12}\cdot2^{20}}=\frac{2^{15}\cdot3^{21}}{2^{26}\cdot3^{12}}=\frac{3^9}{2^{11}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5-x}{2}=\frac{2}{5-x}\)
=> (5-x).(5-x)=2.2=4
(5-x)2=22
=>5-x =2
x=5-2=3
![](https://rs.olm.vn/images/avt/0.png?1311)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(\Leftrightarrow\left(x+5\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=8\\x+5=-8\end{matrix}\right.\Leftrightarrow x\in\left\{3;-13\right\}\)