TH1 \(x\ge0;\left|2x-3\right|=2x-3\)
\(2x-3-5=7x+1\)
\(\Leftrightarrow2x-7x=3+5+1=9\)
\(\Leftrightarrow-5x=9\Rightarrow x=-\frac{9}{5}\left(ktm\right)\)
TH2:\(x< 0;\left|2x-3\right|=-\left(2x-3\right)\)
\(-\left(2x-3\right)-5=7x+1\)
\(\Leftrightarrow-2x+3-5=7x+1\)
\(\Leftrightarrow-2x-7x=-3+5+1=3\)
\(\Leftrightarrow-9x=3\Rightarrow x=-\frac{3}{9}=-\frac{1}{3}\left(tm\right)\)
Vậy \(x=-\frac{1}{3}\)
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