chứng minh x ²+2x+2>0 với mọi x x ²-x+1>0 với mọi x -x ²+4x-5 <0 với mọi x

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1. a) $x^2+2x+2$

   $=x^2+2x.1+1^2+1$

   $=(x+1)^2+1$ 

   Vì $(x+1)^2≥0$

   $→(x+1)^2+1>0∀x$

   b) $x^2-x+1$

   $=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}$

   $=(x-\dfrac{1}{2})^2+\dfrac{3}{4}$

   Vì $(x-\dfrac{1}{2})^2≥0$

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

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

   $=-(x^2-2.2.x+4.1)$

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

   Vì $-(x-2)^2≤0$

   $→-(x-2)^2-1<0∀x$

    

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2. a) $x^2 + 2x + 2$

   $=x^2 + 2x + 1 +1 $

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

   Ta có:

   $(x+1)^2 \geq 0, \, \forall x$

   $\to (x+1)^2 +1 >0\, \forall x$

   b) $x^2 – x + 1$

   $=x^2 – 2.\dfrac{1}{2}x + \dfrac{1}{4} + \dfrac{3}{4}$

   $= \left(x – \dfrac{1}{2}\right)^2 + \dfrac{3}{4}$

   Ta có:

   $ \left(x – \dfrac{1}{2}\right)^2 \geq 0, \,\forall x$

   $\to \left(x – \dfrac{1}{2}\right)^2 + \dfrac{3}{4} >0, \,\forall x$

   c) $-x^2 + 4x – 5$

   $= -x^2 + 4x – 4 – 1$

   $= -(x^2 – 4x + 4) – 1$

   $= -(x-2)^2 – 1$

   Ta có:

   $(x-2)^2 \geq 0, \,\forall x$

   $\to -(x-2)^2 \leq 0,\,\forall x$

   $\to -(x-2)^2 – 1 < 0, \,\forall x$

   Reply