Đáp án: có một câu đề lỗi nhá , là 2x^2 +4x+1
$ x^2 -6x +11 =0$
$⇔x²-6x +9 +2 = 0$
$⇔(x-3)² +2 =0$
$⇔(x-3)² ≥ 0 ∀ x$
$⇔ (x-3)² +2 > 0 ∀ x$
$9x² -6x +2 =0$
$⇔ (3x)² -2.3x .(-1) +1² +1 =0$
$⇔(3x-1)² +1 =0$
$⇔(3x-1)² ≥ 0 ∀ x$
$⇔(3x-1)² +1> 0 ∀ x$
$25x² +30x +17 =0$
$⇔(5x)² + 2 . 5 .3 + 9 + 6 =0$
$⇔ (5x+3)² +6 =0$
$⇔(5x+3)² ≥ 0 ∀ x$
$⇔(5x+3)² +6 >0 ∀ x$
$x^2 -x + \dfrac{1}{2} =0$
$⇔ x^2 - 2 . x . \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} = 0$
$⇔ (x-\dfrac{1}{2})² +\dfrac{1}{4} =0$
$⇔(x-\dfrac{1}{2})² ≥ 0 ∀ x$
$⇔(x -\dfrac{1}{2})² +\dfrac{1}{4} > 0 ∀ x$
$x^2 +5x +8 =0$
$⇔ x^2 . 2 . x .\dfrac{5}{2} + \dfrac{10}{4} + \dfrac{11}{2} =0$
$⇔(x +\dfrac{5}{2})² +\dfrac{11}{2} =0$
$⇔(x+\dfrac{5}{2})² ≥ 0 ∀ x$
$⇔(x+\dfrac{5}{2})² +\dfrac{11}{2} > 0$
$(2x)² +4x +1 = 0$
$⇔ (2x)² + 2x .2 .1 +1 =0$
$⇔(2x+1)²≥ 0 ∀ x$ $\text{( đề lỗi) }$
$3x^2 -12x +16 =0$
$⇔ (\sqrt[]{3}x)² - 2 . 2\sqrt[]{3} .√3x + 12 +4 =0$
$⇔(\sqrt[]{3}x+2\sqrt[]{3})² +4 =0$
$⇔ (\sqrt[]{3}x +2\sqrt[]{3})² ≥ 0 ∀ x$
$⇔ (\sqrt[]{3}x +2\sqrt[]{3})² +4 >0 ∀ x$
$x^4 +x^2 +2 =0$
$⇔ (x^2)^2$$+ 2 . x^2 .\dfrac{1}{2} + \dfrac{1}{4} +\dfrac{7}{4} =0$
$⇔(x^2 + \dfrac{1}{2})^2 +\dfrac{7}{4} =0$
$⇔(x^2 + \dfrac{1}{2})^2 ≥ 0 ∀ x $
$⇔ (x^2 +\dfrac{1}{2})^2 + \dfrac{7}{4} > 0 ∀ x $
