Đáp án: $A=\dfrac{1012}{1011}$
Giải thích các bước giải:
Ta có:
$A=\dfrac{x^3+8x^2+19x+12}{x^3+6x^2+11x+6}$
$\to A=\dfrac{(x^3+x^2)+(7x^2+7x)+(12x+12)}{(x^3+x^2)+(5x^2+5x)+(6x+6)}$
$\to A=\dfrac{(x+1)x^2+7x(x+1)+12(x+1)}{x^2(x+1)+5x(x+1)+6(x+1)}$
$\to A=\dfrac{(x+1)(x^2+7x+12)}{(x+1)(x^2+5x+6)}$
$\to A=\dfrac{x^2+7x+12}{x^2+5x+6}$
$\to A=\dfrac{(x^2+3x)+(4x+12)}{(x^2+3x)+(2x+6)}$
$\to A=\dfrac{x(x+3)+4(x+3)}{x(x+3)+2(x+3)}$
$\to A=\dfrac{(x+3)(x+4)}{(x+3)(x+2)}$
$\to A=\dfrac{x+4}{x+2}$
Với $x=2020$
$\to A=\dfrac{2020+4}{2020+2}=\dfrac{1012}{1011}$
