Giải thích các bước giải:
a.$2x^2-6x+1=9x^2-6x+1-7x^2=(3x-1)^2-(x\sqrt 7)^2=(3x-1-x\sqrt 7)(3x-1+x\sqrt 7)$
b.đkxđ : $x\ne \pm 2$
$\dfrac{x+1}{x-2}=\dfrac{1}{x^2-4}$
$\to\dfrac{x+1}{x-2}=\dfrac{1}{(x-2)(x+2)}$
$\to (x+2)(x+1)=1$
$\to x^2+3x+2=1$
$\to x^2+3x+1=0$
$\to (x+\dfrac 32)^2=\dfrac 54$
$\to x+\dfrac 32=\pm\dfrac{\sqrt{5}}{2}$
$\to x=\dfrac{-3\pm\sqrt{5}}{2}$