[(1/x)-(1/y)+(1/z)]xy/(xy+yz+zx)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 00:04:51
[(1/x)-(1/y)+(1/z)]xy/(xy+yz+zx)

[(1/x)-(1/y)+(1/z)]xy/(xy+yz+zx)
[(1/x)-(1/y)+(1/z)]xy/(xy+yz+zx)

[(1/x)-(1/y)+(1/z)]xy/(xy+yz+zx)
[(1/x)-(1/y)+(1/z)]xy/(xy+yz+zx)=
[(1/x)* xy-(1/y)* xy+(1/z)* xy]/(xy+yz+zx)=
(y - x + xy/z)/(xy+yz+zx)=
(y - x + xy/z) * z/(xy+yz+zx)*z =
( zy - zx + xy ) / z(xy+yz+zx)

已知实数X.Y.Z满足(Y+Z)分之X+(Z+X)分之Y+(X+Y)分之Z=1,则(Y+Z)分之X平方+(Z+X)分之Y平方+(X+Y)分之Z平方的值为( ) 已知x,y,z>0,xyz(x+y+z)=1,求证(x+y)(x+z)>=2 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) 1.设有比例式:x/(y+z)=y/(x+z)=z/(x+y),有比例性质,得x/(y+z)=y/(x+z)=z/(x+y)=(x+y+z)/[(y+z)+(x+z)+(x+y)]=0.5x/(y+z)=y/(x+z)=(x-y)/[(y+z)-(x+z)]=-1,由此可得0.5=-1试分析推理产生错误的原因 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) 已知xyz满足(x/y+z)+(y/z+x)+(z/x+y)=1,则代数式(x^2/y+z)+(y^2/z+x)+z^2/x+y的值为 运用乘法公式计算1:(x+y+2z)(x-y-2z)2:(x+2y-z)² 已知(1)X-Y+Z=0 (2)X+2Y-3Z=0 求X:Y:Z 已知实数xyz满足x/y+z+y/z+x+z/x+y=1求x^2/y+z+y^2/z+x+z^2/x+y的值 纠正下!:已知实数xyz满足(x/y+z)+(y/z+x)+(z/x+y)=1求(x^2/y+z)+(y^2/z+x)+(z^2/x+y)的值 z对x的偏导数=(x^2+y^2)/x ,当x=1,z(x,y)=sin y 求z(x,y) 已知x/2=y/3=z/4,求下列各式 (1)(x+y+z)/x (2)(x-y+2x/(x-y-2z) 已知x+y+z=0,求x(y分之1+z分之)+y(x分之一+z分之一)+z(x分之一+y分之一)的值 已知正数x,y,z满足x+y+z=xyz.求不等式1/(x+y) + 1/(y+z) + 1/(z+x)的最大值 已知x::y:z=3:4:5,(1)求x+y分之z的值;(2)若x+y+z=6,求x,y,z. 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 3(x-1)³y-(1-x)³z m(x-y)²-x+y x²(x+y)(y-x)-xy(x+y)(x-y) 因式分解 把下列各式因式分解 (1)、a^2(x-y)-4a(y-x)+4(x-y) (2)、(x+y)(x-y)-(x+z)(x-z) 已知x ,y ,z都是正数且满足xyz(x+y+z)=1试求(x+y)(y+z)取得最小值时x,y,z的值各是多少?书上的解答是这样的:因为x ,y ,z都是正数,所以(x+y)+(y+z)>(x+z),(y+z)+(z+x)>(x+y),(z+x)+(x+y)>(y+z),于是可