求大师计算啊[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/22 13:30:02
求大师计算啊[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3x²y)²
[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3x²y)²
=[9x²y²*x^4-2x²*9x²y^4*(y/2)]/(9x^4y²
=[x^4(9x²y²)-2x²(9x²y^4)*(y/2)]/(9x^4y²)
=[x^4(9x²y²)-2x²y²(9x²y²)*(y/2)]/(9x²y²)x²
=[(9x²y²)(x^4-2x²y²)*(y/2)]/(9x²y²)x²
=[(x^4-2x²y²)*(y/2)]/x²
=[x²(x²-2y²)*(y/2)]/x²
=(x²-2y²)*(y/2)
=[ ( x² - 2y² ) y ] / 2
对吗
还有更简便的方法吗
[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3x²y)²
=[9x²y²*x^4-2x²*9x²y^4*(y/2)]/(9x^4y²
=[x^4(9x²y²)-2x²(9x²y^4)*(y/2)]/(9x^4y²)
=[x^4(9x²y²)-2x²y²(9x²y²)*(y/2)]/(9x²y²)x²
=[(9x²y²)(x^4-2x²y²)*(y/2)]/(9x²y²)x²
=[(x^4-2x²y²)*(y/2)]/x²
=[x²(x²-2y²)*(y/2)]/x²
=(x²-2y²)*(y/2)
=[ ( x² - 2y² ) y ] / 2
对吗
还有更简便的方法吗
错的.
[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3x²y)²
=[9x²y²*x^4-2x²*9x²y^4*(y/2)]/(9x^4y²)
=(9x^6y²-9x^4y^5)/(9x^4y²)
=(9x^6y²-9x^4y^5)/(9x^4y²)
=9x^4y²(x^4-y^3)/(9x^4y²)
=x^4-y^3
再问: (9x^6y²-9x^4y^5) 这是怎么算的啊?
再答: 同底数幂相乘,底数不变,指数相加。 9x²y²*x^4=9x^6y² 2x²*9x²y^4*(y/2) =x²*9x²y^4*y=9x^4y^5
再问: 玛利亚,神圣玛利亚
[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3x²y)²
=[9x²y²*x^4-2x²*9x²y^4*(y/2)]/(9x^4y²)
=(9x^6y²-9x^4y^5)/(9x^4y²)
=(9x^6y²-9x^4y^5)/(9x^4y²)
=9x^4y²(x^4-y^3)/(9x^4y²)
=x^4-y^3
再问: (9x^6y²-9x^4y^5) 这是怎么算的啊?
再答: 同底数幂相乘,底数不变,指数相加。 9x²y²*x^4=9x^6y² 2x²*9x²y^4*(y/2) =x²*9x²y^4*y=9x^4y^5
再问: 玛利亚,神圣玛利亚
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