谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)

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谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)=ab(x-2y)-a(2y-z

谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)
谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)

谢谢ab(x-2y)-a(2y-z)+ad(2z-4y)
=ab(x-2y)-a(2y-z)+2ad(2y-z)

谢谢ab(x-2y)-a(2y-z)+ad(2z-4y) (y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x+z-2y)的值我已经做到这了:设(X-Y)为a (Z-Y)为b (Z-X)为c则原题= (-a*-c)/-(a+b)(b+c)+ab/-(c+b)(a+c)+c*-b/(-a+c)(a+b) x+y+z=12 x-y+z=2 x-z+y+a=90 x+y+a+z=100,求x y z a (2x-y+z-2c+m)(m+y-2x-2c-z)(a+3b)(a^2-3ab+9b^2)-(a-3b)(a^2+3ab+9b^2)(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z) 合并同类项:3(-ab+2a)-(3a-b)【x-(y-z)】-【x+(y-z)】 已知x+y+z=3a,(a≠0,x,y,z不全等).求(x-a)(y-a)+(y-a)(z-a)+(z-a)(x-a)/(x-a)^2+(y-a)^2+(z-a)^2 x+y+z=3a(a不等于0,x,y,z不全相等)求[(x-a)(y-a)+(y-a)(z-a)+(z-a)(x-a)]/[(x-a)^2+(y-a)^2+(z-a)^2] 已知X+Y+Z=3a求(X-a)(Y-a)+(Y-a)(Z-a)+(Z-a)(X-a)/(X-a)2;+(Y-a)2;+(Z-a)2;,a≠0,X,Y,Z不全相等, 2.化简(1)(2x-y+z-2c+m)(m+y-2x-2c-z)(2)(a+3b)(a^2-3ab+9b^2)-(a-3b)(a^2+3ab+9b^2)(3)(x+y)^2(y+z-x)+(x-y)^2(x+y+z)(x+y-z)卷子明天就交,救命.我只有5分, 分解因式 4a(x-y)²-a²(y-x)³ (x+y+z)(x-y+z)+(y-x+z) (y-x-z)(1)4a(x-y)²-a²(y-x)³ (2)(x+y+z)(x-y+z)+(y-x+z) (y-x-z) (a-b)+(-2a-b)=?(x+y-z)+(x-y+z)-(x-y-z)=? 化简[x-(y-z)]-[(x-y)-z]A.2yB.2zC.-2yD.-2z 初一数学题与代数式有关化简(a+b)(a^2-3ab=9b^2)-(a-3b)(a^2+3ab+9b^2)(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z)已知z^2=x^2+y^2,化简(x+y+z)(x-y+z)(-x+y+z)(x+y-z)已知x(x-1)-(x^2-y)=-2求x^2+y^2/2-xy不需要完整的过程有2步 a²-b²-4ab-4c²=a²-()()+x²-xy+2y²=2x²+xy=3y²(-x+y+z)(x+y-z)=[y-()][y+()] 几道因式分解..4ab(a+b)^2-6a^2b(a+b)(x+y)^2(x-y)+(x+y)(x-y)^22a(a-3)²-6a²(3-a)+8a(a-3)24xy²z²(x+y-z)-32xyz(z-x-y)²+8a(a-3) z=ln(x+a^-y^2) 对y求导, 计算:(a+2)(3-a)+(2x+y+z)(2x-y-z) (x+y+z)^2-(x-y-z)^2(a-b)^n+2 -(a-b)^n