中科新闻网上海市徐汇区党委书记:一道计算题 急用
来源：百度文库 编辑：中科新闻网 时间：2024/05/12 12:09:08
计算：(-1/2x^4y^3-5/3x^3y^4+3/2x^2y^2)÷(-1/3x^2y^2) （要步骤）
(-1/2x^4y^3-5/3x^3y^4+3/2x^2y^2)÷(-1/3x^2y^2)
=(-1/2x^4y^3)÷(-1/3x^2y^2) +(-5/3x^3y^4)÷(-1/3x^2y^2) +(3/2x^2y^2)÷(-1/3x^2y^2)
=3/2*x^2y+5xy^2-9/2
(-1/2x^4y^3-5/3x^3y^4+3/2x^2y^2)÷(-1/3x^2y^2)
=(-1/2x^4y^3)÷(-1/3x^2y^2) +(-5/3x^3y^4)÷(-1/3x^2y^2) +(3/2x^2y^2)÷(-1/3x^2y^2)
=3/2*x^2y+5xy^2-9/2
（-1/2x^4y^3除-1/3x^2y^2)-(5/3x^3y^4除-1/3x^2y^2)+(3/2x^2y^2除-1/3x^2y^2)
=3/2x^2y-5xy^2+(-9/2)
=3/2x^2y-5xy^2-9/2
(-1/2x^4y^3-5/3x^3y^4+3/2x^2y^2)÷(-1/3x^2y^2)
=(-1/2x^4y^3)÷(-1/3x^2y^2) +(-5/3x^3y^4)÷(-1/3x^2y^2) +(3/2x^2y^2)÷(-1/3x^2y^2)
=3/2x^2y+5/xy^2-9/2
前几位结果好象不对吧