Giải thích các bước giải:
$x^2+\dfrac{x^2-2x+10}{(x-1)^2}=2x+7|\dfrac{x^2-2x-2}{x-1}|$
$\rightarrow x^2-2x+1+\dfrac{(x-1)^2+9}{(x-1)^2}-1=7|\dfrac{(x-1)^2-3}{x-1}|$
$\rightarrow (x-1)^2+\dfrac{9}{(x-1)^2}=7|(x-1)-\dfrac{3}{x-1}|$
$\rightarrow (x-1)^2-2.(x-1).\dfrac{3}{x-1}+\dfrac{9}{(x-1)^2}+6=7|(x-1)-\dfrac{3}{x-1}|$
$\rightarrow |(x-1)-\dfrac{3}{x-1}|^2-7|(x-1)-\dfrac{3}{x-1}|+6=0$
$\rightarrow (|(x-1)-\dfrac{3}{x-1}|-6).(|(x-1)-\dfrac{3}{x-1}|-1)=0$
Do $|(x-1)-\dfrac{3}{x-1}|\ge |x-1|+\dfrac{3}{|x-1|}\ge 2\sqrt{ |x-1|.\dfrac{3}{|x-1|}}=2\sqrt{3}>1$
$\rightarrow |(x-1)-\dfrac{3}{x-1}|=6$
$\rightarrow (x-1)-\dfrac{3}{x-1}=6\rightarrow (x-1)^2-6(x-1)-3=0\rightarrow x^2-8x+4=0\rightarrow x=4\pm 2\sqrt{3}$
Hoặc $(x-1)-\dfrac{3}{x-1}=-6\rightarrow (x-1)^2+6(x-1)-3=0\rightarrow x^2+4x-8=0\rightarrow x=-2\pm 2\sqrt{3}$
