Câu 3:
a) $2x^2+x=0⇔x(2x+1)=0⇔$\(\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\x=-1/2\end{array} \right.\)
Vậy x=0 hoặc x=-1/2
b) $2x^3-3x^2+x+a=(2x^3+4x^2)-(7x^2+14x)+(15x+30)-30+a$
$⇔2x^2(x+2)-7x(x+2)+15(x+2)+(a-30)$
Ta thấy để biểu thức chia hết cho (x+2) thì a-30=0⇔a=30
Câu 4:
a) Đk: $x^2+5x\neq0⇔x(x+5)\neq0⇔\left \{ {{x\neq0} \atop {x\neq-5}} \right.$
b)$P=\frac{x^3-25x}{x^2+5x}= \frac{x(x^2-25)}{x(x+5)} =\frac{x(x-5)(x+5)}{x(x+5)}=x-5$
c) $P=\frac{1}{2}⇔x-5=\frac{1}{2}⇔x=\frac{1}{2}+5=\frac{11}{2}$
Câu 5:
$M=(x-1)(x+2)(x+3)(x+6)=[(x-1)(x+6)][(x+2)(x+3)]$
$=(x^2+5x-6)(x^2+5x+6)$
Đặt $t=x^2+5x$
$M=(t-6)(t+6)=t^2-36≥-36$
Dấu = xảy ra $⇔t=0⇔x^2+5x=0⇔x(x+5)=0$\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
