Giải thích các bước giải:
a.Ta có :
$x^2-x-2017.2018=0$
$\to (x^2-2018x)+(2017x-2017.2018)=0$
$\to x(x-2018)+2017(x-2018)=0$
$\to (x+2017)(x-2018)=0$
$\to x\in\{-2017,2018\}$
b.Ta có :
$\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}$
$\to\dfrac{1}{(x+1)(x+3)}+\dfrac{1}{(x+3)(x+5)}+\dfrac{1}{(x+5)(x+7)}=\dfrac{1}{9}$
$\to\dfrac{2}{(x+1)(x+3)}+\dfrac{2}{(x+3)(x+5)}+\dfrac{2}{(x+5)(x+7)}=\dfrac{2}{9}$
$\to\dfrac{x+3-(x+1)}{(x+1)(x+3)}+\dfrac{x+5-(x+3)}{(x+3)(x+5)}+\dfrac{x+7-(x+5)}{(x+5)(x+7)}=\dfrac{2}{9}$
$\to\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}=\dfrac 29$
$\to\dfrac{1}{x+1}-\dfrac{1}{x+7}=\dfrac{2}{9}$
$\to 9\left(x+7\right)-9\left(x+1\right)=2\left(x+1\right)\left(x+7\right)$
$\to 2x^2+16x-40=0$
$\to x^2+8x-20=0$
$\to (x-2)(x+10)=0\to x\in\{2,-10\}$