已知A是整式,且(-1/3xyz)^2·A=1/3 - 知识问答

已知A是整式,且(-1/3xyz)^2·A=1/3x^(2n+1)y^(n+3)z^3÷5x^(2n-1)y^(n+1)

2个回答

∵(-1/3xyz)²·A=1/3x^(2n+1)y^(n+3)z^3÷5x^(2n-1)y^(n+1)
=(1/15)x^[2n+1-(2n-1)]y^[n+3-(n+1)]z³
=(1/15)x²y²z³
∴A=(1/15)x²y²z³÷(-1/3xyz)²
=(1/15)x²y²z³÷[(1/9)x²y²z²]
=(3/5)z
∴A=(3/5)z

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