1.（2x-y+z)^2=2.x^2+y^2=13,xy=6,求x^4+y^4的值.3.因式分解：1）-1/2x^2y^

(2x-y+z)²=4x²+y²+z²-4xy-2yz+4zx
x²+y²=13
两边平方得
x^4+2x²y²+y^4=169
(x^4+y^4)+2(xy)²=169
x^4+y^4+2×6²=169
x^4+y^4=97
-(1/2)x²y³-x²y+2x³y²z=(1/2)x²y(-y²-2+4xyz)=x²y(4xyz-y²-2)/2
(x+y)²(x-y)+(x+y)(y-x)²
=(x+y)²(x-y)+(x+y)(x-y)²
=[(x+y)+(x-y)](x+y)(x-y)
=2x(x+y)(x-y)