解方程组2xy-5√(xy+1)=10 x^2+y^2=34

∵2xy-5√(xy+1)=10
==>2xy+2-5√(xy+1)=12
==>2(xy+1)-5√(xy+1)-12=0
==>[2√(xy+1)+3][√(xy+1)-4]=0 (用十字相乘法分解)
==>√(xy+1)-4=0 (∵2√(xy+1)+3>0)
==>√(xy+1)=4
∴xy=15.(1)
∵x^2+y^2=34
∴x^2+y^2+2xy=64
==>(x+y)²=64
==>x+y=±8
∵当x+y=8时,代入(1)
得x(8-x)=15
==>x²-8X+15=0
==>(x-3)(x-5)=0
==>x1=3,x2=5
∴y1=5,y2=3
∵当x+y=-8时,代入(1)
得x(-8-x)=15
==>x²+8x+15=0
==>(x+3)(x+5)=0
==.x1=-3,x2=-5
∴y1=-5,y2=-3
故原方程有4组(1)x=3,y=5
(2)x=5,y=3
(3)x=-3,y=-5
(4)x=-5,y=-3