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\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left|x-9\right|=8\)
\(\Leftrightarrow x-9=\pm8\)
\(\Leftrightarrow\orbr{\begin{cases}x-9=8\\x-9=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=17\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)
Vậy \(x=17\)hoặc \(x=1\)
b);c) Ta có: \(17-\left|x+5\right|=14\)
\(\Leftrightarrow\left|x+5\right|=3\)
\(\Leftrightarrow x+5=\pm3\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(TM\right)\\x=-8\left(TM\right)\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=-8\)
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a) x = 4
b) x = -8
c) | x | - 9 = -2 + 17
| x | = 15 + 9
| x | = 24
x = 24 hoặc x = -24
d) |x – 9| = -2 + 17
|x – 9| = 15
x – 9 = 15 hoặc x – 9 = -15
x = 24 hoặc x = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x + 9 = 2 - 17
x + 9 = - 15
x = -15 – 9
x = -24
Vậy x = -24
b) x - 17 = (-11) . (-5)
x – 17 = 55
x = 17 + 55
x = 72
c)| x – 5 | = (-4)2
| x – 5 | = 16
x – 5 = 16 hoặc x – 5 = -16
x = 21 hoặc x = -11
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a)\(-\frac{12}{7}\cdot\left(\frac{3}{4}-x\right)\cdot\frac{1}{4}=0\)
=>\(\frac{3}{4}-x=0\)
=>\(x=\frac{3}{4}\)
\(x=\frac{3}{4}\)
b) \(x:\frac{17}{8}=-\frac{2}{5}\cdot\left(-\frac{9}{17}\right)\)
=>\(x:\frac{17}{8}=\frac{18}{85}\)
=>\(x=\frac{18}{85}\cdot\frac{17}{8}=\frac{9}{20}\)
\(x=\frac{9}{20}\)
![](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 ...
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
Ta có:
\(X:\frac{17}{8}=\frac{-2}{5}.\frac{-9}{17}\)
\(X:\frac{17}{8}=\frac{\left(-2\right).\left(-9\right)}{5.17}\)
\(X:\frac{17}{8}=\frac{18}{85}\)
\(\Rightarrow X=\frac{18}{85}.\frac{17}{8}\)
\(X=\frac{9}{20}\)
Vậy \(X=\frac{9}{20}\)