$\frac{x}{x+1}$ +$\frac{3-3x}{x^2-x+1}$ - $\frac{x+4}{x^3+1}$
$\frac{x(x^2-x+1)}{x+1}$ +$\frac{(3-3x)(x+1)}{x^2-x+1}$ - $\frac{x+4}{x^3+1}$
$\frac{x^3-x+1}{x^3+1}$ +$\frac{(3-3x)^2}{x^3+1}$ - $\frac{x+4}{x^3+1}$
$\frac{x^3-x^2+1x^2-x+1+3-3x^3-x-4}{x^3+1}$
$\frac{x^3-4x^2-x}{x^3+1}$
vói a=2 ta có
$\frac{x^3-4x^2-x}{x^3+1}$ =2
$\frac{x^3-4x^2-x}{x^3+1}$ -2=0
$\frac{x^3-4x^2-x}{x^3+1}$ -$\frac{2x^3+2}{x^3+1}$ =0
$\frac{-x^3-4x^2-x+2}{x^3+1}$=0
nếu a >0
=>$\frac{x^3-4x^2-x}{x^3+1}$ >0
=>x^3-4x^2-x>0
=> x^3-4x^2>x
=>x^3>x+4x^2
=> x.x^2>x.(4x+1)
=>x^2>4x+1
=>x^2+3>4.(x+1)
đến đây mình chịu rồi cho mình câu trả lời hay nhất nha
