Bài 1:
a: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)-x\left(x+3\right)=-7x+3\)
\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)
=>0x=0(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
b: \(\Leftrightarrow2x+3< 6-3+4x\)
=>2x+3<4x+3
=>-2x<0
hay x>0