简单计算一下即可，答案如图所示
2xy√Z = x+2y4x²y²Z = (x+2y)²8xy²Z+4x²y²Z'x = 2(x+2y) Z'x = [2(x+2y)-8xy²Z]/4x²y² (1)8x²yZ+4x²y²Z'y = 4(x+2y) Z'y = [4(x+2y)-8xy²Z]/4x²y² (2)还可以把Z的表达式代入(1)、(2)。
