\(3x^3+4x^2+2x+1=0\)
\(\Leftrightarrow\left(3x^3+x^2+x\right)+\left(3x^2+x+1\right)=0\)
\(\Leftrightarrow x\left(3x^2+x+1\right)+1\left(3x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x+1\right)=0\)
Ta có:\(3x^2+x+1=3\left(x^2+x.\frac{1}{3}+\frac{1}{3}\right)\)
\(=3\left(x^2+2.x.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{1}{3}\right)\)
\(=3\left[\left(x+\frac{1}{6}\right)^2+\frac{11}{36}\right]\ge3.\frac{11}{36}=\frac{11}{12}>0\forall x\)
Do đó x + 1 = 0 tức là x = -1
\(3x^3+3x^2+x^2+x+x+1=0\)
\(3x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right).\left(3x^2+x+1\right)=0\)
+)\(3x^2+x+1=0\Leftrightarrow3.\left(x^2+x+\frac{1}{3}\right)=0\Leftrightarrow3.\left(x+\frac{1}{6}\right)^2+\frac{11}{12}=0\left(loai\right)\)
+) x+1=0 <=> x=-1
