a, $M=x^2+3x-4$
$M=x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}$
$M=(x+\dfrac{3}{2})^2-\dfrac{25}{4}$
$(x+\dfrac{3}{2})^2≥0∀x$
$⇔(x+\dfrac{3}{2})^2-\dfrac{25}{4}≥-\dfrac{25}{4}∀x$
Dấu $=$ xảy ra khi
$x+\dfrac{3}{2}=0$
$⇔x=\dfrac{-3}{2}$
Vậy $M_{min}=\dfrac{-25}{4}⇔x=\dfrac{3}{2}$
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b, $N=3x^2-2x-4$
$N=3x^2-2.\sqrt{3}x.\dfrac{\sqrt{3}}{3}+\dfrac{1}{3}-\dfrac{13}{3}$
$N=(\sqrt{3}x-\dfrac{\sqrt{3}}{3})^2-\dfrac{13}{3}$
$(\sqrt{3}x-\dfrac{\sqrt{3}}{3})^2≥0∀$
$⇔(\sqrt{3}x-\dfrac{\sqrt{3}}{3})^2-\dfrac{13}{3}≥\dfrac{-13}{3}∀x$
Dấu $=$ xảy ra khi
$\sqrt{3}x-\dfrac{\sqrt{3}}{3}=0$
$⇔\sqrt{3}x=\dfrac{\sqrt{3}}{3}$
$⇔x=\dfrac{1}{3}$
Vậy $N_{min}=\dfrac{-13}{3}⇔x=\dfrac{1}{3}$
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c, $P=x^2+y^2-2x-4y+8$
$P=x^2-2x+1+y^2-4y+4+3$
$P=(x-1)^2+(y-2)^2+3$
$(x-1)^2≥∀x,(y-2)^2≥0∀y$
$⇔(x-1)^2+(y-2)^2+3≥3∀x,y$
Dấu $"="$ xảy ra khi
$+,x-1=0$
$⇔x=1$
$+,y-2=0$
$⇔y=2$
Vậy $P_{min}=3⇔x=1;y=2$
