\(\left(3x+2\right)-\left(x-1\right)=49x+1\)
\(\Leftrightarrow3x+1-x+1-49x-1=0\)
\(\Leftrightarrow-47x-1=0\)
\(\Leftrightarrow-47x=1\)
\(\Leftrightarrow x=-\dfrac{1}{47}\)
Ta có : (3x+2) - (x-1) = 49x + 1
\(\Leftrightarrow\) 3x + 2 - x +1 = 49x + 1
\(\Leftrightarrow\) 2x + 3 = 49x +1
\(\Leftrightarrow\) 49x + 1 - 2x -3 = 0
\(\Leftrightarrow\) 47x - 2 = 0
\(\Leftrightarrow\) x = \(\dfrac{2}{47}\)
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