Đáp án:
a) $S_{}$ = {$-3;_{}$ -$\frac{5}{2}$}
b) $S_{}$ = {$1;-2_{}$}
Giải thích các bước giải:
a) $2x(x + 3) + 5(x + 3) = 0_{}$
⇔ $2x^{2}+6x+5x+15=0$
⇔ $2x^{2}+11x+15=0$
⇔ $x^{2}+11x+15.2=0$
⇔ $x^{2}+11x+30=0$
⇔ $x^2+5x+6x+30=0_{}$
⇔ $x(x+5)+6(x+5)=0_{}$
⇔ $(x+6)(x+5)=0_{}$
⇔ \(\left[ \begin{array}{l}x+6=0\\x+5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-6\\x=-5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x= -\frac{6}{2} \\x= -\frac{5}{2} \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-3\\x=-\frac{5}{2}\end{array} \right.\)
Vậy: $S_{}$ = {$-3;_{}$ -$\frac{5}{2}$}
b) $3x(x – 1) + 6( x –1) = 0_{}$
⇔ $3x^{2}-3x+6x-6=0$
⇔ $3x^{2}+3x-6=0$
⇔ $x^{2}+3x-6.3=0$
⇔ $x^{2}+3x-18=0$
⇔ $x^{2}+6x-3x-18=0$
⇔ $x(x+6)-3(x+6)=0_{}$
⇔ $(x-3)(x+6)=0_{}$
⇔ \(\left[ \begin{array}{l}x-3=0\\x+6=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=-6\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{3}{3} \\x=-\frac{6}{3} \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy: $S_{}$ = {$1;-2_{}$}
