计算,【1】[x(x^2-2x+3)-3x]÷1/2x^2 【2】[(2xy-3)(【1】[x(x^2-2x+3)-3x]÷1/2x^2【2】[(2xy-3)(2xy+3)+(xy+3^2]÷xy

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/06 07:53:12
计算,【1】[x(x^2-2x+3)-3x]÷1/2x^2【2】[(2xy-3)(【1】[x(x^2-2x+3)-3x]÷1/2x^2【2】[(2xy-3)(2xy+3)+(xy+3^2]÷xy计算,

计算,【1】[x(x^2-2x+3)-3x]÷1/2x^2 【2】[(2xy-3)(【1】[x(x^2-2x+3)-3x]÷1/2x^2【2】[(2xy-3)(2xy+3)+(xy+3^2]÷xy
计算,【1】[x(x^2-2x+3)-3x]÷1/2x^2 【2】[(2xy-3)(
【1】[x(x^2-2x+3)-3x]÷1/2x^2
【2】[(2xy-3)(2xy+3)+(xy+3^2]÷xy

计算,【1】[x(x^2-2x+3)-3x]÷1/2x^2 【2】[(2xy-3)(【1】[x(x^2-2x+3)-3x]÷1/2x^2【2】[(2xy-3)(2xy+3)+(xy+3^2]÷xy