x 2=y 3=z 5,2x 3y-z=16
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x=log2(y)则X1+2X2+3X3=log2(y1)+2log2(y2)+3log2(y3)=log2(y1)+log2(y2^2)+log2(y3^3)=log2(y1y2^2y3^3)=1所
设x2=y3=z5=k,则x=2k,y=3k,z=5k,∴6k+6k-5k=14,k=2,∴x=4,y=6,z=10.答:x,y,z的值分别为4,6,10.
x3y+2x2y2+xy3=xy(x2+2xy+y2)=xy(x+y)2,∵x+y=5,∴(x+y)2=25,x2+y2+2xy=25,∵x2+y2=13,∴xy=6,∴xy(x+y)2=6×25=1
因为:X3-Y3-Z3=3XYZ所以:X3+(-Y)3+(-Z)3-3X(-Y)(-Z)=0(X-Y-Z)(X2+Y2+Z2+XY+XZ-YZ)=0所以:1.X-Y-Z=02.X2+Y2+Z2+XY+
反应前XY均为0价,反应后化合价有变化,四氧化还原反应.提一句,4X2+Y2=X3Y+Y2去掉Y2的话是4X2=X3Y,这是不可能的,元素本身发生了变化,应该是核反应
由(x+y+z)2-(x2+y2+z2)可得xy+xz+yz=-5x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)可得xyz=14再问:谢谢,我看一下其他的答案在采纳再
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
∵x+y=4,∴(x+y)2=16,∴x2+y2+2xy=16,而x2+y2=14,∴xy=1,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=14-2=12.
∵z=12+32i=cosπ3+isinπ3,z+2z2+3z3+4z4+5z5+6z6=cosπ3+isinπ3+2cos2π3+2sin2π3i+3cosπ+3sinπi+4cos4π3+4sin
设x2=y3=z4=k,则x=2k,y=3k,z=4k,∵2x-3y+4z=22,∴4k-9k+16k=22,∴k=2,∴x+y-z=2k+3k-4k=k=2.
如果你的X2是x的平方,X3是x的三次方那么答案是:-(x-y+z)*(x-y-z)*(x+y-z)
y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得:y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以
(2x4-4x3y-x2y2)-2(x4-2x3y-y3)+x2y2=2x4-4x3y-x2y2-2x4+4x3y+2y3+x2y2=2y3,因为化简的结果中不含x,所以原式的值与x值无关.
第一题,证明:x2+y3+z5=t7,取x=3,y=1,z=1,t=2,3*2+1*3+1*5=2*7,等式成立,设m为任意正整数,当x=3*m,y=1*m,z=z*m,t=2*m时,有3*m*2+1
x1/y1=x2/y2=x3/y3=1/2y1=2x1,y2=2x2,y3=2x3(x1+x2-x3)/(y1+y2-y3)=﹙x1+x2-x3)/[2﹙x1+x2-x3)]=½
设x2=y3=z5=t,则x=2t,y=3t,z=5t,∴x+3y−zx−3y+z=2t+9t−5t2t−9t+5t=-3.故答案是:-3.再问:=v=O(∩_∩)O谢谢
(x1x2x3)=(y1y2y3)A===>(y1y2y3)=(x1x2x3)A^{-1}是逆矩阵,不是转置.再问:请问为什么呢,为什么这么做,意义何在再答:有点儿抽象。有时候,一些实际问题需要考虑反
根据√x/y+√y/z+√z/xx,y,z应全>0或全0将题中两式相减得:x^3-x^2+y^3-y^2+z^3-z^2=1(x-1)x^2+(y-1)y^2+(z-1)z^2=1因为x,y,z>0,