ĐK:....... pt (1) ⇔x−1+√(x−1)3+1=3√y+2+√(3√y+2)3+1
⇔(x−1−3√y+2)(1+1√...+√...)=0
⇔x−1=3√y+2
pt (2) Tt
3√x−1−√x2−6x+6=x−1+1
⇔x−1+1+√(x−1)2−4(x−1)+1=3√x−1 (3)
+) x=1 k tm hệ pt
+0 x≠1 chia cho √x−1
(3) ⇔√x−1+1√x−1+√x−1−4+1x−1=3
đặt t=√x−1+1√x−1>2
⇔t+√t2−6=3
⇔t=52
⇔x=5 or x=54
⇒(x;y)=(5;62);(54;−12764)