1) -x^2 – 1 2) -(x + 1)^2 3) -(x + 1)^2 – 3 4) -x^2 – 2x – 1 5) -x^2 – 2x – 4 6) -x^2 – 4x – 1 7) -x^2 – x – 1 8) -2x^2 + 6x – 8 9) -2x^2 – 3x – 4 10)

0 bình luận về “1) -x^2 – 1 2) -(x + 1)^2 3) -(x + 1)^2 – 3 4) -x^2 – 2x – 1 5) -x^2 – 2x – 4 6) -x^2 – 4x – 1 7) -x^2 – x – 1 8) -2x^2 + 6x – 8 9) -2x^2 – 3x – 4 10)”

1. Đáp án:

   $ 1) -x^2-1$

   Vì $-x^2 ≤ 0 $

   Nên $ -x^2-1 < 0 ∀ x$

   $2) -(x+1)^2$

   Vì $(x+1)^2 ≥ 0 ∀ x$

   Nên $-(x+1)^2 ≤ ∀ x$

   $3) -(x+1)^2 – 3$

   Vì $-(x+1)^2 ≤ 0 ∀ x$

   Nên $-(x+1)^2 -3 < 0∀x$

   $4) -x^2-2x-1$

   $ = -(x^2+2x+1)$

   $ = -(x+1)^2$

   Vì $(x+1)^2 ≥ 0 ∀ x$

   Nên $-(x+1)^2 ≤ 0 ∀ x$

   $5) -x^2-2x-4$

   $ = -(x^2+2x+4)$

   $=-(x^2 +2.x.1+ 1+3)$

   $ = -(x+1)^2 -3$

   Vì $-(x+1)^2 ≤ 0 ∀ x$

   Nên $ -(x+1)^2 -3 < 0 ∀ x$

   6) sai đề ko chứng minh đc

   $7) -x^2-x-1$

   $ = -(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 ∀ x$

   Nên $ -(x+\dfrac{1}{2})^2 -\dfrac{3}{4} < 0 ∀ x$

   $8) -2x^2+6x-8$

   $ = -(2x^2-6x+8)$

   $ = -[(√2x)^2- 2 . √2x. \dfrac{3\sqrt[]{2}}{2} + \dfrac{9}{2} + \dfrac{7}{2}]$

   $ = -(√2x -\dfrac{3\sqrt[]{2}}{2})^2 – \dfrac{7}{2}$

   Vì $ -(√2x -\dfrac{3\sqrt[]{2}}{2})^2 ≤ 0 ∀ x$

   Nên $ -(√2x- \dfrac{3\sqrt[]{2}}{2})^2 – \dfrac{7}{2}$ < 0 ∀ x$

   $9) -2x^2 -3x -4$

   $ = -(2x^2 +3x +4)$

   $ = -[(√2x)^2 + 2 . √2x . \dfrac{3\sqrt[]{2}}{4}+ \dfrac{9}{8} + \dfrac{23}{8}]$

   $ = -(√2x +\dfrac{3\sqrt[]{2}}{4})^2 -\dfrac{23}{8}$

   Vì $ -(√2x +\dfrac{3\sqrt[]{2}}{4})^2 ≤ 0 ∀ x$

   Nên $ -(√2x +\dfrac{3\sqrt[]{2}}{4})^2  – \dfrac{23}{8} < 0 ∀ x$

   $10) -2x^2 + 2x – 4$

   $ = -(2x^2-2x+4)$

   $ = -[(√2x)^2 – 2 . √2x . \dfrac{\sqrt[]{2}}{2} + \dfrac{1}{4} + \dfrac{15}{4}]$

   $ = -(√2x -\dfrac{\sqrt[]{2}}{2})^2 – \dfrac{15}{4}$

   Vì $ -(√2x -\dfrac{\sqrt[]{2}}{2})^2 ≤ 0 ∀ x$

   Nên $ -(√2x -\dfrac{\sqrt[]{2}}{2})^2 – \dfrac{15}{4} < 0 ∀ x $

    

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