\(\left(x+2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
=>\(x^3+6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
=>\(15x^2+12x+7=3x^2+3x+1\)
=>\(12x^2+9x+6=0\)
\(\Leftrightarrow4x^2+3x+2=0\)
\(\text{Δ}=3^2-4\cdot4\cdot2=9-32=-23< 0\)
=>Phương trình vô nghiệm
\(\left(x+2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x-1\right)^3\)
\(x^3+6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
\(x^3+15x^2+12x+7-x^3-3x^2-3x-1=0\)
\(12x^2+9x+6=0\)
\(4x^2+3x+2=0\)
\(4x^2+2.2x.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)
\(\left(2x+\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)
Do \(\left(2x+\dfrac{3}{4}\right)^2\ge0\)
\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2+\dfrac{23}{16}>0\)
\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\) là vô lý
Vậy \(S=\varnothing\)
