求大师计算啊[(-3xy)²*x^4-2x²(3xy²)²*(y/2)]/(-3x²y)²
问题描述：
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
答
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(9x^6y²-9x^4y^5) 这是怎么算的啊?同底数幂相乘，底数不变，指数相加。9x²y²*x^4=9x^6y²2x²*9x²y^4*(y/2) =x²*9x²y^4*y=9x^4y^5玛利亚,神圣玛利亚