Đáp án + Giải thích các bước giải:
Bài 2:
$a) 4(2x-3)=2x(2x-3)\\⇔4(2x-3)-2x(2x-3)=0\\⇔(2x-3)(4-2x)=0$
$⇔\left[ \begin{array}{l}2x-3=0\\4-2x=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=2\end{array} \right.$
$b)2x(x-5)=3(5-x)\\⇔2x(x-5)+3(x-5)=0\\⇔(x-5)(2x+3)=0$
$⇔\left[ \begin{array}{l}x-5=0\\2x+3=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=5\\x=-\dfrac{3}{2}\end{array} \right.$
c$)9x^{2}-16=0\\⇔(3x)^2-4^2\\⇔(3x-4)(3x+4)=0$
$⇔\left[ \begin{array}{l}3x-4=0\\3x+4=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{array} \right.$
$d)x^{3}-3\sqrt{3}x^{2}+9x-3\sqrt{3}=0\\⇔x^{3}-3.x^{2}.\sqrt{3}+3.x.3-(\sqrt{3})^{3}=0\\⇔(x-\sqrt{3})^{3}=0\\⇔x-\sqrt{3}=0\\⇔x=\sqrt{3}$
Bài 3:
$A=-2x^2-6x\\=-2(x^{2}+3x)\\=-2(x^{2}+3x+\dfrac{9}{4}-\dfrac{9}{4})\\=-2(x+\dfrac{3}{2})^{2}+\dfrac{9}{2}$
Ta có: `-2(x+(3)/(2))^2<=0`
`⇔-2(x+3/2)^2+9/2<=9/2`
Vậy `max_A=9/2` khi `x=-3/2`
