\(a.6x-3=5x+2\)
\(\Leftrightarrow6x-3-5-2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
\(S=\left\{1\right\}\)
\(b.2-3x=5x-6\)
\(\Leftrightarrow2-3x-5x+6=0\)
\(\Leftrightarrow-8x+8=0\)
\(\Leftrightarrow x=1\)
\(S=\left\{1\right\}\)
\(c.\left|3x\right|=2x+7\left(1\right)\)
\(TH_1:3x\ge0\Leftrightarrow x\ge0\)
\(\left(1\right)\Leftrightarrow3x=2x+7\)
\(\Leftrightarrow3x-2x=7\)
\(\Leftrightarrow x=7\left(n\right)\)
\(TH_2:3x< 0\Leftrightarrow x< 0\)
\(\left(1\right)\Leftrightarrow-3x=2x+7\)
\(\Leftrightarrow-3x-2x=7\)
\(\Leftrightarrow-5x=7\)
\(\Leftrightarrow x=\dfrac{-5}{7}\left(n\right)\)
Vậy pt (1) có tập n0 S = \(\left\{7,\dfrac{-5}{7}\right\}\)
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