Đáp án:
$b) x^2 -8x +20$
$= x^2 - 2 . x . 4 + 16 +4$
$=(x-4)^2 +4$
Vì $(x-4)^2 ≥ 0$
Nên $(x-4)^2 + 4 > 0 ∀ x$
$c) x^2 + x + \dfrac{1}{2}$
$= x^2 + 2. x . \dfrac{1}{2} + \dfrac{1}{4} +\dfrac{1}{4}$
$= (x+\dfrac{1}{2})^2 + \dfrac{1}{4}$
Vì $(x+\dfrac{1}{2})^2 ≥ 0$
Nên $(x+\dfrac{1}{2})^2 + \dfrac{1}{4} > 0 ∀ x$
$d) 2x^2 +4x +5$
$= (\sqrt[]{2}x)^2 + 2. \sqrt[]{2}x . \sqrt[]{2} + 2 +3$
$=(\sqrt[]{2}x + \sqrt[]{2})^2 + 3$
Vì $(\sqrt[]{2}x + \sqrt[]{2})^2 ≥ 0$
Nên $(\sqrt[]{2}x +\sqrt[]{2})^2 + 3 > 0 ∀ x$
$e) 3x^2 -6x+7$
$ = (\sqrt[]{3}x)^2 - 2 . \sqrt[]{3}x . \sqrt[]{3} +3 +4$
$ = (\sqrt[]{3}x - \sqrt[]{3})^2 +4$
Vì $(\sqrt[]{3}x - \sqrt[]{3})^2 ≥ 0$
Nên $(\sqrt[]{3}x - \sqrt[]{3})^2 +4 >0 ∀ x$
$g) x^2 -2x+y^2 +4y+6$
$ = (x^2 -2 .x +1) + (y^2 +2 .y .2 +4) +1$
$ = (x-1)^2 + (y +2)^2 +1$
Vì $(x-1)^2 + (y+2)^2 ≥ 0$
Nên $(x-1)^2 + (y+2)^2 +1 >0 ∀ x$
$h) 4x^2 -12x + 11$
$= (2x)^2 - 2 . 2x .3 + 9 +2$
$ = (2x -3)^2+2$
Vì $(2x-3)^2 ≥ 0$
Nên $(2x-3)^2 + 2 > 0 ∀ x$
