Đặt:√x=t⇒pt⇔√28−3t2=7t3−12t2−14t+24⇔√28−3t2−(6−t)=7t3−12t2−14t+24−(6−t)$\Leftrightarrow \frac{-4(t-1)(t-2)}{\sqrt{28-3t^2}+6-t}=(t-20(t-1)(7t-9)$$\Leftrightarrow t=1-hoặc- t=2\Leftrightarrow x=1;4$
Đặt
:√x=t⇒pt⇔√28−3t2=7t3−12t2−14t+24⇔√28−3t2−(6−t)=7t3−12t2−14t+24−(6−t)$\Leftrightarrow \frac{-4(t-1)(t-2)}{\sqrt{28-3t^2}+6-t}=(t-2
)(t-1)(7t-9)$$\Leftrightarrow t=1-hoặc- t=2\Leftrightarrow x=1;4$