a) Ta có: 12-5x=37
\(\Leftrightarrow5x=-25\)
hay x=-5
Vậy: x=-5
b) Ta có: 7-3|x-2|=-11
\(\Leftrightarrow3\left|x-2\right|=18\)
\(\Leftrightarrow\left|x-2\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-4\right\}\)
c) Ta có: \(x+\dfrac{2}{8}=-\dfrac{15}{4}\)
\(\Leftrightarrow x=\dfrac{-15}{4}-\dfrac{2}{8}=\dfrac{-15}{4}-\dfrac{1}{4}\)
hay x=-4
Vậy: x=-4
a, \(\Leftrightarrow5x=12-37=-25\)
\(\Leftrightarrow x=-\dfrac{25}{5}=-5\)
Vậy ...
b, \(\Leftrightarrow3\left|x-2\right|=7+11=18\)
\(\Leftrightarrow\left|x-2\right|=\dfrac{18}{3}=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
Vậy ...
c, \(\Leftrightarrow x=-\dfrac{15}{4}-\dfrac{2}{8}=-4\)
Vậy ..
a)-5; b)8 hoặc -4;c)-4 nhé. Bn tham khảo nhé
Giải:
a) \(12-5x=37\)
\(5x=12-37\)
\(5x=-25\)
\(x=-5\)
b) \(7-3.\left|x-2\right|=-11\)
\(3.\left|x-2\right|=7-\left(-11\right)\)
\(3.\left|x-2\right|=18\)
\(\left|x-2\right|=18:3\)
\(\left|x-2\right|=6\)
\(\Rightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
c) \(x+\dfrac{2}{8}=\dfrac{-15}{4}\)
\(x=\dfrac{-15}{4}-\dfrac{2}{8}\)
\(x=-4\)
