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a) Ta có:
\(x^2-2x+4\)
\(=x^2-2x+1+3\)
\(=\left(x-1\right)^2+3\ge3>0\forall x\)
b) Ta có:
\(\left(x-2\right)\left(x-4\right)+3\)
\(=x^2-4x-2x+8+3\)
\(=x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x-3\right)^2+2\ge2>0\forall x\)
c) \(-x^2+4x-5\)
\(=-x^2+4x-4-1\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)
d) \(-2x^2+5x-19\)
\(=-2\left(x^2-\dfrac{5}{2}x+\dfrac{19}{2}\right)\)
\(=-2\left(x^2-2\cdot\dfrac{5}{4}\cdot x+\dfrac{25}{16}+\dfrac{127}{16}\right)\)
\(=-2\left(x-\dfrac{5}{4}\right)^2-\dfrac{127}{8}\le-\dfrac{127}{8}< 0\forall x\)
a. Ta có:
$x^2-2x+4$
$=x^2-2x+1+3$
$=(x-1)^2+3$
Ta thấy: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+3\ge3>0\forall x\)
hay biểu thức $x^2-2x+4$ luôn dương với mọi $x$.
$---$
b. Ta có:
$(x-2)(x-4)+3$
$=x^2-6x+8+3$
$=x^2-2\cdot x\cdot 3+3^2+2$
$=(x-3)^2+2$
Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2>0\forall x\)
hay biểu thức $(x-2)(x-4)+3$ luôn dương với mọi $x$.
$---$
c. Ta có:
$-x^2+4x-5$
$=-(x^2-4x+4)+4-5$
$=-(x^2-2\cdot x\cdot 2+2^2)-1$
$=-(x-2)^2-1$
Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-1\le-1< 0\forall x\)
hay biểu thức $-x^2+4x-5$ luôn âm với mọi $x$.
$---$
d. Ta có:
$-2x^2+5x-19$
$=-2\Bigg(x^2-\dfrac{5}{2}x\Bigg)-19$
$=-2\Bigg[x^2-2\cdot x \cdot \dfrac54 +\Bigg(\dfrac54\Bigg)^2 \Bigg] +2\cdot \Bigg(\dfrac54\Bigg)^2 -19 $
$=-2\Bigg(x-\dfrac54)^2 -\dfrac{127}{8}$
Ta thấy: \(\left(x-\dfrac{5}{4}\right)^2\ge0\forall x\)
\(\Rightarrow-2\left(x-\dfrac{5}{4}\right)^2\le0\forall x\)
\(\Rightarrow-2\left(x-\dfrac{5}{4}\right)^2-\dfrac{127}{8}\le-\dfrac{127}{8}< 0\forall x\)
hay biểu thức $-2x^2+5x-19$ luôn âm với mọi $x$.
$Toru$
