计算下列几道题★1.（2x+y-5z)(2x+y+5z)2.16(y+3)^2(y-3)^2-(2y+1)^2(2y-1

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/11/07 20:37:29

计算下列几道题★
1.（2x+y-5z)(2x+y+5z)
2.16(y+3)^2(y-3)^2-(2y+1)^2(2y-1)^2
3.(1/2x-1/3y)(1/2x+1/3y)(1/4x^2+1/9y^2)

1.（2x+y-5z)(2x+y+5z)=(2x+y)^2-(5z)^2=4x^2+y^2+4xy-25z^2
2.16(y+3)^2(y-3)^2-(2y+1)^2(2y-1)^2=[4(y^2-9)]^2-(4y^2-1)^2
=(4y^2-36)^2-(4y^2-1)^2=(4y^2-36+4y^2-1)(4y^2-36-4y^2+1)
=(8y^2-37)(-35)=1295-280y^2
3.(1/2x-1/3y)(1/2x+1/3y)(1/4x^2+1/9y^2)
=(1/4x^2-1/9y^2)(1/4x^2+1/9y^2)
=1/16x^4-1/81y^4
主要是应用平方差公式

1.（x+y-z)+(x-y+z)-(x-y-z) 2.2x-(3x-2y+3)-(5y-2) 计算下列几道题★1.（2x+y-5z)(2x+y+5z)2.16(y+3)^2(y-3)^2-(2y+1)^2(2y-1 x/2=y/3=z/5 x+3y-z/x-3y+z {x+y+z=1;x+3y+7z=-1;z+5y+8z=-2 因式分解：25x y^2 z^2 (x+y-z)-30xyz(z-x-y)^2+5x y z^3 (z-x-y) (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai x+y+z=2 x-3y+2z=1 2x+2y+z+5 要具体步骤.. 3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) (y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z {5x-3y+z=2{5x+2y-4z=3{-5x+y-z=2 {x-y-z=-1{3x+5y+7z{4x-y+2x=- x-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 分别求出x=?y=?z=?