## 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

Question

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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8 tháng 2020-11-02T04:55:09+00:00 2 Answers 84 views 0

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 $-(x-2)^2≤0$

$→-(x-2)^2-1<0∀x$

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:

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

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

$\to -(x-2)^2 – 1 < 0, \,\forall x$