Lời giải:
\(x^3+6x^2+12x+35=0\)
\(\Leftrightarrow x^3+5x^2+x^2+5x+7x+35=0\)
\(\Leftrightarrow x^2(x+5)+x(x+5)+7(x+5)=0\)
\(\Leftrightarrow (x+5)(x^2+x+7)=0\)
\(\Rightarrow \left[\begin{matrix} x+5=0\\ x^2+x+7=0\end{matrix}\right.\Leftrightarrow \Rightarrow \left[\begin{matrix} x=-5\\ (x+\frac{1}{2})^2=-\frac{27}{4}< 0(\text{vô lý})\end{matrix}\right.\)
Vậy PT có nghiệm duy nhất $x=-5$
\(x^3+6x^2+12x+35=0\)
\(\Leftrightarrow x^3+5x^2+x^2+5x+7x+35=0\)
\(\Leftrightarrow x^2\left(x+5\right)+x\left(x+5\right)+7\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+x+7\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+x+\frac{1}{4}+\frac{27}{4}\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{27}{4}\right]=0\)
\(\Leftrightarrow x+5=0\Leftrightarrow x=-5\)
Vậy...
