提高拓展-整式的化简VIP免费

1.已知x+y=3，xy=1，求x2+y2与(x-y)2的值.2.已知a²-3a+1=0，求(a+1)²-5a的值.x2+y2=(x+y)2-2xy=9-2=7(x-y)2=(x+y)2-4xy=9-4=5原式=a2+2a+1-5a=a2-3a+1=03.已知x2+y2-4x-6y+13=0，求x-y的值.x²-4x+4+y2-6y+9=0(x-2)2+(y-3)2=0∴x=2，y=3，x-y=-1