高考问答连云港新闻港城365最新消息:一道数学题 急用
来源：百度文库 编辑：高考问答 时间：2024/04/27 23:39:25
用平方差来计算下面的题目：
(2x+7)(2x-7)-4x(x-5)
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
(2x+7)(2x-7)-4x(x-5)
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
(2x+7)(2x-7)-4x(x-5)
=4x2-49-4x2+20
=-29
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
=x2-1/9y2-9x2+4y2
=35/9y2-8x2
(2x+7)(2x-7)-4x(x-5)
=4*x^2-49-4*x^2+20
=-29
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
=x^2-y^2/9-9*x^2+4*y^2
=-8*x^2+35*y^2/9
(2x+7)(2x-7)-4x(x-5)
=4*x^2-49-4*x^2+20
=-29
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
=x^2-1/9*y^2-9*x^2+4*y^2
=-8*x^2+35/9*y^2
(2x+7)(2x-7)-4x(x-5)
=4*x^2-49-4*x^2+20x
=20x-49
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
=x^2-1/9*y^2-9*x^2+4*y^2
=35/9*y^2 -8*x^2
(2x+7)(2x-7)-4x(x-5)=4*x^2-49-4*x^2+20=-29
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)=x^2-1/9*y^2-9*x^2+4*y^2=-8*x^2+35/9*y^2
2x+7)(2x-7)-4x(x-5)
=4x2-49-4x2+20
=-29
(x-1/3y)(x+1/3y)-(3x-2y)(3x+2y)
=x2-1/9y2-9x2+4y2
=35/9y2-8x2