Giải thích các bước giải:
Bài 2:
$\dfrac{x^4-4x^2+1}{x^3+x}+\dfrac{x^4+3x^2+1}{x^2}=4$
$\to \dfrac{x^4-4x^2+1}{x(x^2+1)}+\dfrac{x^4+3x^2+1}{x^2}=4$
$\to x\left(x^4-4x^2+1\right)+\left(x^4+3x^2+1\right)\left(x^2+1\right)=4x^2\left(x^2+1\right)$
$\to x^5-4x^3+x+x^6+4x^4+4x^2+1=4x^4+4x^2$
$\to x^6+x^5-4x^3+x+1=0$
$\to \left(x-1\right)^2\left(x^4+3x^3+5x^2+3x+1\right)=0$
$\to x=1$
Vì $x^4+3x^3+5x^2+3x+1=x^2(x^2+3x+\dfrac 94)+(\dfrac{11}4x^2+3x+1)=x^2(x+\dfrac 32)^2+\dfrac{11}4(x+\dfrac{6}{11})^2+\dfrac{2}{11}>0$
