Bài 3 :
a) 2(x + 3) - $x^2$ - 3x = 0
2(x + 3) - x(x + 3) = 0
(2 - x)(x + 3) = 0
TH1 : 2 - x = 0 => x = 2
TH2 : x + 3 = 0 => x = -3
b) $x^2$ - 10x = -25
$x^2$ - 10x + 25 = 0
$(x - 5)^2$ = 0
x - 5 = 0 => x = 5
Bài 4 :
$x^2$ - 4x + 7
= $x^2$ - 4x + 4 + 3
= $(x - 2)^2$ + 3
$(x - 2)^2$ + 3 ≥ 3 ∀x
=> $(x - 2)^2$ + 3 > 0 ∀x
Bài 6 :
a) ($2x^3 + 5x^2 - 2x + 3$) : ($2x^2$ - x + 1)
$2x^3 - x^2 + x$ | x + 3
______________________|
$6x^2 - 3x + 3$ |
$6x^2 - 3x + 3$|
_____________________________
0
*khó trình bày quá
