Đáp án:
$ 1.A = x^2+2x+2$
$ = (x+1)^2 +1 $
Vì $(x+1)^2 ≥ 0$
Nên$ (x+1)^2 +1 0 ∀ x$
$2 . B = 4x^2 +4x +3 $
$ = (2x +1)^2 +2 $
Vì $(2x+1)^2 ≥ 0$
Nên $(2x+1)^2 +2 > 0∀x$
$3. C =x^2 -x +1 $
$ = x^2 - 2 . x . \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}$
$ = (x-\dfrac{1}{2})^2 + \dfrac{3}{4}$
Vì $(x -\dfrac{1}{2})^2 ≥ 0 $
Nên $(x-\dfrac{1}{2})^2 + \dfrac{3}{4} > 0 ∀x$
$4 . D =3x^2 -6x +4$
$ = (\sqrt[]{3}x)^2 - 2 . \sqrt[]{3}x. \sqrt[]{3} + 3 + 1$
$ = (\sqrt[]{3}x - \sqrt[]{3})^2 + 1$
Vì$ (\sqrt[]{3}x -\sqrt[]{3})^2 ≥ 0$
Nên $(\sqrt[]{3}x-\sqrt[]{3})^2 + 1 > 0 ∀x$
$ 5 . E = 4x^2 +3x + 2$
$ = (2x)^2 + 2 . 2x . \dfrac{3}{4} + \dfrac{9}{16} + \dfrac{23}{16}$
$ = (2x + \dfrac{3}{4})^2 + \dfrac{23}{16}$
Vì$ (2x + \dfrac{3}{4})^2 ≥ 0$
Nên$2x + \dfrac{3}{4})^2 + \dfrac{23}{16} > 0 ∀ x $
$6 P = x^2 -6x +10 $
$ = (x-3)^2 +1 $
Vì $(x-3)^2 ≥ 0$
Nên$(x-3)^2 +1 > 0 ∀ x $
$ 7 . Q = x^2 +x + 1$
$ = x^2 + 2 . x . \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}$
$ = (x+\dfrac{1}{2})^2 + \dfrac{3}{4}$
Vì $(x +\dfrac{1}{2})^2≥0$
Nên$ ( x + \dfrac{1}{2})^2 + \dfrac{3}{4} > 0 ∀ x$
$8 .E=(x-3).(x-5) +4 $
$ = x^2 -5x -3x +15 +4$
$ = x^2 -8x +19$
$ = x^2 - 2 . x . 4 + 16 +3$
$ =(x-4)^2 +3$
Vì $(x-4) ≥ 0$
Nên $(x-4)^2 +3 > 0 ∀ x$
