Trl:
\(3\left|x-1\right|-5=7\)
\(\Rightarrow3\left|x-1\right|=7+5\)
\(\Rightarrow3\left|x-1\right|=12\)
\(\Rightarrow\left|x-1\right|=12:3\)
\(\Rightarrow\left|x-1\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=4+1\\x=-4+1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
Vậy \(x\in\left\{5;-3\right\}\)
\(3\left|x-1\right|-5=7\)
\(\Leftrightarrow3\left|x-1\right|=2\)
\(\Leftrightarrow\left|x-1\right|=\frac{2}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{2}{3}\\x-1=-\frac{2}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{3}\end{cases}}\)
Vậy....
Nhầm
\(3\left|x-1\right|-5=7\)
\(\Leftrightarrow3\left|x-1\right|=12\)
\(\Leftrightarrow\left|x-1\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
Vậy...
\(3\left|x-1\right|-5=7\)
\(\Rightarrow3\left|x-1\right|=7+5=12\)
\(\Rightarrow\left|x-1\right|=12:3=4\)
\(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
Vậy \(x\in\left\{5;-3\right\}\)
3|x-1|-5=7
-> 3|x-1| = 7+5
-> 3|x-1|= 12
-> |x-1|= 12:3 = 4
TH1: x-1 = 4 TH2: x-1 = -4
<=> x= 4+1 = 5 <=> x= -4 +1 = -3
Vậy: x 
