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,

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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,
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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,
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,

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,
(1)[(2x+z-y)+(m-2c)]*[(m-2c)-(2x+z-y)]=(m-2c)平方—(2x+z-y)平方
=(m-2c)^2-[(x+z)+(x-y)]^2=(m-2c)^2-(x+z)^2-2(x+z)(x-y)-(x-y)^2
(2)=(a+3b)*[(a-3b)^2+3ab]-(a-3b)*[(a+3b)^2-3ab]
=(a+3b)*(a-3b)^2+3ab(a+3b)-(a-3b)*(a+3b)^2+3ab(a-3b)
=[(a+3b)*(a-3b)]*(a-3b)-[(a+3b)*(a-3b)]*(a+3b)+6a^2b
=(a^2-9b^2)*(-6b)+6a^2b=54b^3

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, 1已知-2m+2n=5,那么5(m-2n)平方+6n-3m-60的值为多少2.若3x+2y+4z=4,x-y+z=2,求x+4y+2z的值3.已知a-b分之x=b-c分之y=c-a分之z,求2(x+y+z)的值 1.m(x-y-z)+n(y+z-x)=(x-y-z)*?2.x^2-(a+1)x+a+?3.6x^2+7x+2=? 数学题(x-y)/(y-z),(y-z)/(z-x),(z-x)/(x-y)已知有理数X,Y,Z两两不相等,则(x-y)/(y-z),(y-z)/(z-x),(z-x)/(x-y)中负数的个数是()A.1个 B.2个 C.3个 D.0个或2个 x.y.z.m都是有理数,并且x+y+2z=m,x+2y+3z=m,那么y与z( ) 已知xyz≥0,x+y+z=1,化简x(2y-z)/(1+x+3y)+y(2z-x) /(1+y+3z) +z(2x-y)/(1+z+3x) 计算下面的分式题.急,我是初一的,1、(y-z)(z-x)/[(x+z-2y)(x+y-2z)]+(z-y)(x-z)/[(x+y-2z)(y+z-2x)+(x-z)(y-z)]/[(y+z-2x)(x+z-2y)]2、若a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b-c+d)的值.3、若x/(x^2-mx+1)=1,求x^3/(x^6-m^3x^3+1)的值.4、如果 1、下列各项是多项式4x²-(y-z²)的一个因式的是() A.4x-y+z B.4x-y-z C.2x-y+z D.2x-y-z 如果X+Y——Z,的反应是按照,1X+M——N,2N+Y——Z+M的过程进行的,则作为催化剂的是如果X+Y——Z,的反应是按照:1.X+M——N2.N+Y——Z+M的过程进行的,则作为催化剂的是()A.Y B.Z C.M D.N下列实 已知A={(x,y)|x=n,y=an+b,n属于Z},B={(x,y)|x=m,y=3m方+15,m属于Z};C={(x,y)|x方+y方小于等于144}问是否存在实数a,b,使得(1)A交B不等于空集;(2)(a,b)属于C同时成立 已知复数Z=2m+(m^+1)i (m∈R) 1.求复数z的对应点Z(x,y)的轨迹C的方程2.若直线y=x+b和上述轨迹C交于不同两点A.B 求AB中点的轨迹方程 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z) 下列算式中与x-y-z的值不相等的是( )A.x-(y-z) B.x-(y+z) C.(x-y)+(-2) D.(-y)+(x-z)长方形的一边长为3m+2n,另一边比它长m-n,则这个长方形的周长为( )A.4m+n B.7m+3n C.-x+y-z D.x-y-z 1题:6m^2(a-2)-9m(2-a)^22题:a(x-y)(y-z)-b(y-x)(z-y) 有关因式分解的.下列从左边到右边的变形,是因式分解的是( )A.(3-x)(3+x)=9-x^2 B.m^3-n^3=(m-n)(m^2+mn+n^2)C.(y+1)(y+3)=―(3-y)(y+1) D.4yz-2y^2z=2y(2z-yz)+z西西. #define M(x,y,z) x*y+z main() { int a=1,b=2,c=3; printf(“%d ”,M(a+b,b+c,c+a)); }#define M(x,y,z) x*y+zmain(){ int a=1,b=2,c=3;printf(“%d ”,M(a+b,b+c,c+a));} 请问怎么计算的? 若a+b+c=1,求√(3a+1)+√(3b+1)+√(3c+1)的最大值过程:设x=√(3a+1),y=√(3b+1),z=√(3c+1),t=x+y+za+b+c=1所以x^2+y^2+z^2=6x^2+y^2=6-z^2设m=x+y+z则x+y=m-z因为x^2+y^2>=(x+y)^2/2所以6-z^2>=(m-z)^2/2所以3z^2-2mz+m^2-12=0m 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答