Rút gọn A ta được:
$A=1-t+\frac{t^2(t+1)}{t^2+t+1},$ với $t=\sqrt{x}\geq 0$
Theo giả thiết thì $A=\frac{6-\sqrt{6}}{5}=1+\frac{1-\sqrt{6}}{5}=1+\frac{t^3+t^2-t^3-t^2-t}{t^2+t+1}$
$\Leftrightarrow \frac{t}{t^2+t+1}=\frac{\sqrt{6}-1}{5}$
$\Leftrightarrow (\sqrt{6}-1)t^2+(\sqrt{6}-6)t+\sqrt{6}-1=0$
$\Leftrightarrow t=\frac{\sqrt{6}\pm \sqrt{2}}{2}\Rightarrow x=(\frac{\sqrt{6}\pm \sqrt{2}}{2})^2=2\pm \sqrt{3}$
x2−5y+3+6y2−7x+4−−−−−−−−−−√=0y(y