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\(-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\)
\(\Rightarrow-26-x+7=0\\ \Rightarrow x+19=0\Rightarrow x=-19\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-26-(x-7)=0 \(\Rightarrow\chi-7=-26\)
\(\Rightarrow\chi=-26+7\)
\(\Rightarrow\chi=-19\)
HTDT
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3-\left(17-x\right)=-12\\ \Rightarrow17-x=15\\ \Rightarrow x=2\\ b,-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\\ c,25+\left(-2+x\right)=5\\ \Rightarrow-2+x=20\\ \Rightarrow x=18\\ d,30+\left(32-x\right)=10\\ \Rightarrow32-x=-20\\ \Rightarrow x=52\)
a) `3-(17-x)=-12`
`3-17+x=-12`
`x=-12-3+17`
`x=2`
b) `-26-(x-7)=0`
`-26-x+7=0`
`-19-x=0`
`x=-19`
c) `25+(-2+x)=5`
`25-2+x=5`
`x=5-25+2`
`x=-18`
d) `30+(32-x)=10`
`30+32-x=10`
`62-x=10`
`x=52`
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(x\left(x+7\right)=0\) \(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b/ \(\left(-x+5\right)\left(3-x\right)=0\) \(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
c/ \(\left|x-1\right|=3\) \(\Rightarrow\left[\begin{matrix}x-1=3\\1-x=3\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
d/ \(-13\left|x\right|=-26\) \(\Rightarrow\left|x\right|=2\) \(\Rightarrow x=\pm2\)
e/ \(x.x-8=-2.\left(-13\right)-\left(-2\right)\)
\(\Rightarrow x^2=36\) \(\Rightarrow\left|x\right|=6\) \(\Rightarrow x=\pm6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
*) \(-7+2x=-37-\left(-26\right)\)
⇔ \(-7+2x=-11\)
⇔ \(2x=\left(-11\right)-\left(-7\right)\)
⇔ \(2x=-4\)
⇔ \(x=\left(-4\right):2\)
⇔ \(x=-2\)
\(\text{(3x +9).(11-x ) =0}\)
⇔\(\left[{}\begin{matrix}3x+9=0\\11=x=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}3x=-9\\x=11\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=-3\\x=11\end{matrix}\right.\)
Vậy x= -3 và x= 11
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5\left(x-7\right)+2018^0=26\)
\(5x-35+1=26\)
\(5x-35=25\)
\(5x=60\)
\(x=12\)
\(5\left(x-7\right)+2018^0=26\)
\(\Rightarrow\left(5x-35\right)+1=26\)
\(\Rightarrow5x-35=25\)
\(\Rightarrow5x=60\)
\(\Rightarrow x=12\)
Vậy x = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
a, -15 + \(x\) = -14 - (-57)
-15 + \(x\) = -14 + 57
\(x\) = -14 + 57 + 15
\(x\) = 58
b, (-14) + \(x\) - 7 = -10
\(x\) = -10 + 7 + 14
\(x\) = -3 + 14
\(x\) = 11
-26 - ( x - 7 ) = 0
x - 7 = -26 - 0
x - 7 = -26
x = -26 + 7
x = -19
k cho tớ nha
@ Bắp gửi tus ạ
-26 - ( x - 7 ) = 0
x - 7 = -26 - 0
x - 7 = -26
x = -26 + 7
x = -19