pt(1) :
(x+y−1)√x+y−1+2(3x+y)=20pt(2) : (3x+y−2)√3x+y−2+2(x+y)=18
đặt a=√x+y−1=>a2=x+y−1=>a2+1=x+y
b=√3x+y−2=>b2=3x+y−2=>b2+2=3x+y
ta có hệ ms : {a3+2b2=16b3+2a2=16
lấy pt(1)-pt(2): a3−b3−2(a2−b2)=0
<=> (a−b)(a2+ab+b2−2(a+b))=0
=>a=b
thế lại x,y rùi giải ra