\(\left|x-1\right|=3x+2\)
Xét các trường hợp:
TH1: x - 1 = 3x + 2
x - 3x = 2 + 1
-2x = 3
x = \(\frac{-3}{2}\)
TH2: 1 - x = 3x + 2
-x - 3x = 2 - 1
-4x = 1
x = \(\frac{-1}{4}\)
Vậy x \(\in\left\{\frac{-3}{2};\frac{-1}{4}\right\}\)
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\(\left|5x-4\right|=\left|x+2\right|\)
\(\Rightarrow\left|5x-4\right|-\left|x+2\right|=0\)
\(\Rightarrow\left(5x-4\right)-\left(x+2\right)=0\)
\(\Rightarrow5x-4-x-2=0\)
\(\Rightarrow4x-6=0\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\frac{3}{2}\)
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