Giải thích các bước giải:
a.Ta có:
$A=\dfrac{x^4+x^3-x^2-2x-2}{x^4+2x^3-x^2-4x-2}$
$\to A=\dfrac{(x^4+2x^3-x^2-4x-2)-(x^3-2x)}{x^4+2x^3-x^2-4x-2}$
$\to A=1-\dfrac{x^3-2x}{x^4+2x^3-x^2-4x-2}$
$\to A=1-\dfrac{x^3-2x}{(x^4+2x^3+x^2)-(2x^2+4x+2)}$
$\to A=1-\dfrac{x(x^2-2)}{x^2(x^2+2x+1)-2(x^2+2x+1)}$
$\to A=1-\dfrac{x(x^2-2)}{(x^2-2)(x^2+2x+1)}$
$\to A=1-\dfrac{x}{x^2+2x+1}$
$\to A=\dfrac{x^2+2x+1-x}{x^2+2x+1}$
$\to A=\dfrac{x^2+x+1}{x^2+2x+1}$
b.Ta có:
$A-\dfrac34=\dfrac{x^2+x+1}{x^2+2x+1}-\dfrac34$
$\to A-\dfrac34=\dfrac{4(x^2+x+1)-3(x^2+2x+1)}{4(x^2+2x+1)}$
$\to A-\dfrac34=\dfrac{x^2-2x+1}{4(x^2+2x+1)}$
$\to A-\dfrac34=\dfrac{(x-1)^2}{4(x+1)^2}\ge 0$
$\to A\ge \dfrac34$
Dấu = xảy ra khi $x-1=0\to x=1$
