a) x-4=(4x+1)(x+2)
\(\Leftrightarrow x-4=4x^2+9x+2\)
\(\Leftrightarrow x-4-4x^2-9x-2=0\)
\(\Leftrightarrow-4x^2-8x-6=0\)
\(\Leftrightarrow-4\left(x^2+2x+\frac{3}{2}\right)=0\)
\(\Leftrightarrow x^2+2x+1+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x+1\right)^2=-\frac{1}{2}\)(vô lý)
Vậy: \(S=\varnothing\)
c) Ta có: |x-1|=x+5(*)
Trường hợp 1: \(x\ge1\)
(*)\(\Leftrightarrow x-1=x+5\)
\(\Leftrightarrow-6=0\)(vô lý)
Trường hợp 2: x<1
(*)\(\Leftrightarrow1-x=x+5\)
\(\Leftrightarrow1-x-x-5=0\)
\(\Leftrightarrow-2x-4=0\)
\(\Leftrightarrow-2x=4\)
hay x=-2(tm)
Vậy: S={-2}
d) Ta có: 3x-1=2
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}