∵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=...