$19.V=(2x-1)^{2}+(x-5)^{2}+4\ge4$
Dấu $"="$ xảy ra
$\Leftrightarrow\begin{cases} 2x-1=0\\x-5=0 \end{cases}$
$\Leftrightarrow\begin{cases} x=\dfrac{1}{2}\\x=5 \end{cases}$
Vậy $MinV=4$ khi $x=\dfrac{1}{2}$ và $x=5$
$20.W=x(x+1)+(x+1)(x+2)+(x+2)(x+3)$
$W=x^{2}+x+x^{2}+2x+x+2+x^{2}+3x+2x+6$
$W=3x^{2}+9x+8$
$W=3(x^{2}+3x+\dfrac{8}{3})$
$W=3(x^{2}+2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{5}{12})$
$W=3(x+\dfrac{3}{2})^{2}+\dfrac{5}{4}\ge\dfrac{5}{4}$
Dấu $"="$ xảy ra
$\Leftrightarrow x+\dfrac{3}{2}=0$
$\Leftrightarrow x=-\dfrac{3}{2}$
Vậy $MinW=\dfrac{5}{4}$ khi $x=-\dfrac{3}{2}$
