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\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
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a)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
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3 . ( 2x - 1 ) - 2 = 13
3 . ( 2x - 1 ) = 12 + 3
3 . ( 2x - 1 ) = 15
2x - 1 = 15 : 3
2x - 1 = 5
2x = 5 + 1 = 6
x = 6 : 2 = 3
Vậy x = 3
\(3\left(2x-1\right)-2=13\)
\(3\left(2x-1\right)=15\)
\(2x-1=5\)
\(2x=6\)
\(x=3\)
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1, \(\left(y+7\right)\left(y-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y+7=0\\y-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-7\\y=5\end{matrix}\right.\)
2, \(25-\left(30+x\right)=\left(-24+3\right)\)
\(\Rightarrow25-\left(30+x\right)=-21\)
\(\Rightarrow30+x=25-\left(-21\right)\)
\(\Rightarrow30+x=25+21\)
\(\Rightarrow30+x=46\)
\(\Rightarrow x=46-30\)
\(\Rightarrow x=16\)
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Lời giải:
** Bổ sung điều kiện $x$ là số nguyên.
a. $24\vdots 2x-1$
$\Rightarrow 2x-1$ là ước của $24$. Mà $2x-1$ lẻ nên $2x-1\in\left\{\pm 1; \pm 3\right\}$
$\Rightarrow x\in \left\{1; 0; 2; -1\right\}$
b.
$x+15\vdots x+6$
$\Rightarrow (x+6)+9\vdots x+6$
$\Rightarrow 9\vdots x+6$
$\Rightarrow x+6\in \left\{\pm 1; \pm 3; \pm 9\right\}$
$\Rightarrow x\in \left\{-7; -5; -3; -9; -15; 3\right\}$
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`a) 3 / 5 x - 1 / 2 = 1 / 7`
`3 / 5 x = 1 / 7 + 1 / 2`
`3 / 5 x = 9 / 14`
`x = 9 / 14 : 3 / 5`
`x = 15 / 14`
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`b) 3 / 5 x + 2 / 7 = 4 / 5`
`3 / 5 x = 4 / 5 - 2 / 7`
`3 / 5 x = 18 / 35`
`x = 18 / 35 : 3 / 5`
`x = 6 / 7`
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\(\dfrac{x-1}{2}=\dfrac{2-x}{3}\)
\(\Rightarrow3\cdot\left(x-1\right)=2\cdot\left(2-x\right)\)
\(\Rightarrow3x-3=4-2x\)
\(\Rightarrow3x+2x=4+3\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\dfrac{7}{5}\)
\(x-\dfrac{1}{2}=2-\dfrac{x}{3}\)
\(x+\dfrac{x}{3}=2+\dfrac{1}{2}\)
\(x\left(1+\dfrac{1}{3}\right)=\dfrac{5}{2}\)
\(x\times\dfrac{4}{3}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}:\dfrac{4}{3}\)
\(x=\dfrac{15}{8}\)