d, \(\left|2x+1\right|=\left|x-1\right|\)
TH1 : 2x + 1 = x - 1 <=> x = -2
TH2 : 2x + 1 = 1 - x <=> 3x = 0 <=> x = 0
f, | x - 1 | - 1 = 50 - 2 | x - 1 |
<=> 3 | x - 1 | = 51 <=> | x - 1 | = 17
TH1 : x - 1 = 17 <=> x = 18
TH2 : x - 1 = -17 <=> x = -16
d) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
f) Ta có: \(\left|x-1\right|-1=50-2\left|x-1\right|\)
\(\Leftrightarrow3\left|x-1\right|=51\)
\(\Leftrightarrow\left|x-1\right|=17\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=17\\x-1=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-16\end{matrix}\right.\)
