Đáp án:
Giải thích các bước giải:
$a, (x+2)(x-2)-(x+1)(x-3)+2$
$=x^2-4-(x^2-3x+x-3)+2$
$=x^2-4-x^2+2x+3+2$
$=2x+1$
$b,4(x-1)(x+1)-5x(x-2+x^2)$
$=4(x^2-1)-5x^2+10x-5x^3$
$=4x^2-4-5x^2+10x-5x^3$
$=-x^2-4+10x-5x^3$
$c,(2x-1)(x+3)-(x-2)(x+5)$
$=2x^2+6x-x-3-(x^2+5x-2x-10)$
$=2x^2+5x-3-x^2-3x+10$
$=x^2+2x+7$
$A=(2n+1)(n^2-3n-1)-2n^3+1$
$=2n^3-6n^2-2n+n^2-3n-1-2n^3+1$
$=-5n^2-5n$
$=-5(n+1)$
$-5 \vdots 5 ⇒ -5(n+1) \vdots 5∀n$
Vậy $A\vdots 5 ∀n$
$(x+3)(x-2)-(x+1)(x-5)$
$=x^2-2x+3x-6-(x^2-5x+x-5)$
$=x^2+x-6-x^2+4x+5$
$=5x-1$
