若|x+y-2|+[2xy-1] 2【這個2是平方】=0，求x2+6xy+y2的值

∵|x+y-2|+[2xy-1]²；=0 |x+y-2|≥0 [2xy-1]²；≥0
∴2xy-1=0 2xy=1
x+y-2=0（x+y）²；=2²；x²；+2xy+y²；=4 x²；+y²；+1=4 x²；+y²；=3
x²；+6xy+y²；=3+1×3=6