ĐK:...(1)⇔(1−y)(√x−y−1)+(x−y−1)=(x−y−1)√y
⇔(x−y−1)(1−y√x−y+1+1−√y)=0
⇔(x−y−1)(1−y)(1√x−y+1+1√y+1)=0
⇔x=y+1ory=1(do(...)>0)
*)y=1:
(2)tt:9−3x=0⇔x=3
*)x=y+1
(2)tt:2y2+3y−2=√1−y
⇔2(y2+y−1)=√1−y−y
⇔(y2+y−1)(1√1−y+y+2)=0
⇔y2+y−1=0⇔y=−1+√52(t/m đk)⇒x=1+√52
KL:....