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![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(12+\left(4-x\right)=-5\)
\(\Leftrightarrow12+4-x=-5\)
\(\Leftrightarrow-x=-5-12-4=-21\)
\(\Leftrightarrow x=21\)
2) \(\left|x-6\right|=5\)
\(\Leftrightarrow\left[\begin{matrix}x-6=5\\x-6=-5\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=11\\x=1\end{matrix}\right.\)
3) \(\left|x-3\right|=4\)
\(\Leftrightarrow\left[\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
4) \(x=-11\)
a, 12+(4-x)=-5
4-x=7
=>x=3
b,|x-6|=5
=>x-6=\(\pm6\)
Xét 2 TH:
\(\Rightarrow\left[\begin{matrix}x-6=6\\x-6=-6\end{matrix}\right. \)
\(\Rightarrow\left[\begin{matrix}x=12\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;12\right\}\)
c, |x-3|=4
x-3=±4
Xét 2 TH:
\(\Rightarrow\left[\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{-1;7\right\}\)
d, 12+11+10+...+x=12
\(\Rightarrow11+10+...+x=0\)
Gọi n là số số hạng của vế trái.
\(\Rightarrow\frac{\left(11+x\right).n}{2}\)=11+10+...+x=0
=>11+x=0
=>x=-11
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
2. \(|x| +|x-1| ≤ 5 \\ \Leftrightarrow |x| + |x-1| ≤ \dfrac{5}{2}\)
\(-∞\) | \(0\) | \(1\) | \(+∞\) | |
\(|x|\) | \(-x\) | \(x\) | \(x\) | \(x\) |
\(|x-1|\) | \(1-x\) | \(1-x\) | \(x-1\) | \(x-1\) |
\(|x|+|x-1|\) | \(1-2x\) | \(1\) | \(2x-1\) | \(2x-1\) |
TH1: \(1-2x ≤ \dfrac{5}{2} \Leftrightarrow x ≥ \dfrac{-3}{4}\)
TH2: \(2x-1 ≤ \dfrac{5}{2} \Leftrightarrow x ≤ \dfrac{7}{4}\)
Vậy....