a: =>1+3x-6=7-x
=>3x-5=7-x
=>4x=12
=>x=3(nhận)
b: \(\Leftrightarrow\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}\)
=>\(x^3-3x^2-x^2+3x-x^3-3x^2=-7x^2+3x\)
=>\(-7x^2+3x=-7x^2+3x\)
=>0x=0(luôn đúng)
Vậy: S=R\{3;-3}
c: =>x(x+2)+(2x-1)(x+1)=0
=>2x^2+2x-x-1+x^2+2x=0
=>3x^2+3x-1=0
\(x=\dfrac{-3\pm\sqrt{21}}{6}\)
d: =>2(x-2)-x-1=3x-11
=>3x-11=2x-4-x-1=x-5
=>2x=6
=>x=3(nhận)