Đáp án:

$a)minx=\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b)max=\dfrac{9}{4} \Leftrightarrow x=\dfrac{1}{2}\\ c)min= -3  \Leftrightarrow x=2\\ d)min= 10 \Leftrightarrow x=-\dfrac{1}{2}\\ e)min= -2= \Leftrightarrow x=1\\ f)min=1 \Leftrightarrow \left\{\begin{array}{l} x=1\\ y=2\end{array} \right.\\ g)min= -1 h=\dfrac{-3\pm \sqrt{5}}{2}$

Giải thích các bước giải:

$a)x^2+x+1\\ =x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge \dfrac{3}{4} \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow x+\dfrac{1}{2}=0 \Leftrightarrow x=-\dfrac{1}{2}$

$b)2+x-x^2\\ =-x^2+x+2\\ =-x^2+x-\dfrac{1}{4}+\dfrac{9}{4}\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4} \le \dfrac{9}{4}  \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{1}{2}=0 \Leftrightarrow x=\dfrac{1}{2}$

$c)x^2-4x+1\\ =x^2-4x+4-3\\ =(x-2)^2-3 \ge -3 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow x-2=0 \Leftrightarrow x=2$

$d)4x^2+4x+11\\ =4x^2+4x+1+10\\ =(2x+1)^2+10 \ge 10 \ \forall \ x $

Dấu "=" xảy ra $\Leftrightarrow 2x+1=0 \Leftrightarrow x=-\dfrac{1}{2}$

$e)3x^2-6x+1\\ =3x^2-6x+3-2\\ =3(x^2-2x+1)-2\\ =3(x-1)^2-2 \ge -2 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow x-1=0 \Leftrightarrow x=1$

$f)x^2-2x+y^2-4y+6\\ =x^2-2x+1+y^2-4y+4+1\\ =(x-1)^2+(y-2)^2+1 \ge 1 \ \forall \ x,y$

Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} x-1=0\\ y-2=0\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} x=1\\ y=2\end{array} \right.$

$g)h(h+1)(h+2)(h+3)\\ =h(h+3)(h+1)(h+2)\\ =(h^2+3h)(h^2+3h+2)(1)\\ h^2+3h=t\\ (1) \Leftrightarrow t(t+2)\\ =t^2+2t\\ =t^2+2t+1-1\\ =(t+1)^2 -1 \ge -1 \ \forall \ x$

Dấu "=" xảy ra

$\Leftrightarrow t+1=0 \Leftrightarrow t=-1\Leftrightarrow h^2+3h=-1\Leftrightarrow h^2+3h+1=0\Leftrightarrow h=\dfrac{-3\pm \sqrt{5}}{2}$