Question

giải pt:

\(\left(x+6\right)\left(\dfrac{x^2+3}{2}-1\right)=0\)

Answer 1

$(x+6)\left(\frac{x^2+3}{2}-1\right)=0$

$\Leftrightarrow \left[\begin{array}{} x+6=0\\ \frac{x^2+3}{2}-1=0 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=-6\\ \frac{x^2+3}{2}=1 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=-6\\ x^2+3=2 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=-6\\ x^2=-1\text{ (vô lí)} \end{array} \right.$

$\Rightarrow x=-6$

#$\mathtt{Toru}$