Giải thích các bước giải:
a.Ta có :
$2x^4-9x^3+14x^2-9x+2=0\to x\ne 0$
$\to 2x^2-9x+14-\dfrac9x+\dfrac{2}{x^2}=0$
$\to 2(x^2+\dfrac{1}{x^2})-9(x+\dfrac{1}{x})+14=0$
$\to 2(x^2+2+\dfrac{1}{x^2})-9(x+\dfrac{1}{x})+10=0$
$\to 2(x+\dfrac{1}{x})^2-9(x+\dfrac{1}{x})+10=0$
$\to (2(x+\dfrac{1}{x})-5)(x+\dfrac{1}{x}-2)=0$
$\to 2(x+\dfrac{1}{x})-5=0\to x\in\{2,\dfrac 12\}$
Hoặc $x+\dfrac{1}{x}-2=0\to x=1$
b.Ta có :
$6x^4+25x^3+12x^2-25x+6=0\to x\ne 0$
$\to 6x^2+25x+12-\dfrac{25}{x}+\dfrac{6}{x^2}=0$
$\to 6(x^2+\dfrac{1}{x^2})+25(x-\dfrac{1}{x})+12=0$
$\to 6(x^2-2+\dfrac{1}{x^2})+25(x-\dfrac{1}{x})+24=0$
$\to 6(x-\dfrac{1}{x})^2+25(x-\dfrac{1}{x})+24=0$
$\to (2(x-\dfrac{1}{x})+3)(3(x-\dfrac{1}{x})+8)=0$
$\to 2(x-\dfrac{1}{x})+3=0\to x\in\{\dfrac 12,-2\}$
Hoặc $3(x-\dfrac{1}{x})+8=0\to x\in\{\dfrac 13,-3\}$
