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{√x+1+√y−1=4√x+6+√y+4=6 ⇔ {x+y+2√(x+1)(y−1)=16x+y+10+2√(x+6)(y+4)=36 ⇔ {2√(x+1)(y−1)=16−x−y(1)2√(x+6)(y+4)−10=16−x−y(2) Lấy (1) trừ (2) theo vế, ta được :
√(x+6)(y+4) = √(x+1)(y−1) + 5
⇔ (x+6)(y+4)=(x+1)(y−1)+25+10√(x+1)(y−1)
⇔ xy+4x+6y+24=xy−x+y+24+10√(x+1)(y−1)
⇔ x+y=2√(x+1)(y−1)
⇔ x2+y2+2xy=4xy−4x+4y−4
⇔ x2+y2−2xy+4x−4y+4=0
⇔ x2−2x(y−2)+(y−2)2=0
⇔ (x−y+2)2=0
⇔ x=y−2(3)
Thế (3) vào (1) ta có
(1)⇔√y−1=2⇔y=5⇒x=3
Vậy (x;y)=(3;5)
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