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Ta có :
\(\frac{-6}{12}=\frac{x}{8}=\frac{-7}{y}=\frac{z}{-18}=\frac{-1}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=\frac{-1}{2}\Rightarrow x=\left(-4\right)\\\frac{-7}{y}=\frac{-1}{2}\Rightarrow y=14\\\frac{z}{-18}=\frac{-1}{2}\Rightarrow z=9\end{cases}}\)
Vậy ...
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\(a,23-x=19-\left(-21\right)\)
\(23-x=19+21\)
\(23-x=40\)
\(x=23-40\)
\(x=-17\)
Vậy \(x\in\left\{-17\right\}\)
\(b,x-7=\left(-2\right)-8\)
\(x-7=\left(-2\right)+\left(-8\right)\)
\(x-7=-10\)
\(x=-10+7\)
\(x=-10-\left(-7\right)\)
\(x=-3\)
Vậy \(x\in\left\{-3\right\}\)
a)23-x=19-(-12)
=>23-x=31
x=23-31
x=-8
Vậy x=-8
b)x-7=(-2)-8
x-7=-10
x =(-10)+7
x =-3
Vậy x=-3
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![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a, 7-x=8-(-7)
=> 7-x=15
=> x= 7-15
=> x=-8
b, x-8=-3+(-8)
=> x-8=-11
=> x=-11+8
=> x=-3
-8+|x|=-12
|x|=-12+8
|x|=-4 mà |x|>=0(với mọi x)
nên \(x\in\phi\)