x=3,y=-1/3,求3x05[2xy²-2(xy-3/2x²y)]要过程._数学解答

x=3,y=-1/3,求3x05[2xy²-2(xy-3/2x²y)]要过程.

人气：415 ℃　时间：2020-01-24 07:22:53

解答

原式=x²(2xy²-2xy+3x²y)
=2x³y²-2x³y+3x⁴y
=2*27*1/9-2*27*(-1/3)+3*81*(-1/3)
=6+18-81
=-57