设z=xy分之一则阿尔法y分之阿尔法z等于什么
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设x/2=y/3=z/4=a则:x=2a;y=3a;z=4a代入得:(xy+yz+zx)/(x^2+y^2+z^2)=(6a^2+12a^2+8a^2)/(4a^2+9a^2+16a^2)=26a^2
∵1/x+1/y=1/6,1/y+1/z=1/9,1/z+1/x=1/15∴(1/x+1/y)+(1/y+1/z)+(1/z+1/x)=1/6+1/9+1/152(1/x+1/y+1/z)=15/90
设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则(1)x+y-z=kz(2)x-y+z=ky(3)-x+y+z=kx(1)+(2)+(3)得x+y+z=k(x+y+z)∴k=1时,
xy/(x+y)=51/x+1/y=1/5yz/(y+z)=7/21/y+1/z=2/7zx/(z+x)=41/x+1/z=1/4(xy+yz+zx)分之xyz=1/(1/x+1/y+1/z)=280
令x/3=y/4=z/5=kx=3ky=4kz=5k原式=(3k)^2+(4k)^2+(5k)^2/3k*4k+4k*5k+3k*5k=50k^2/47k^2=50/47
x/z=y/3=-2/5y=-6/5(xy+yz+zx)/z^2=(-5/6*x/z-5/6+x/z)/z=(1/3-5/6-2/5)/z=(-9/10)/z=-9/(10z)
x+y分之xy=1,y+z分之yz=2,z+x分之zx=3每个等式左右均取倒数,所以:1/x+1/y=11/y+1/z=1/21/z+1/x=1/3设:1/x=a1/y=b1/z=ca+b=1----
令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0
令3分之x=4分之y=6分之z=kx=3k,y=4kz=6k(xy+yz+xz)/(x^2+y^2+z^2)=k^2(12+18+24)/k^2(36+16+9)=54/61
A♁B={0,4,5}(A♁B)♁C={0,8,10}所有元素之和为:0+8+10=18
由已知得(x+y)/(xy)=1(y+z)/(yz)=1/2(z+x)/(zx)=1/3变形:1/x+1/y=1(1)1/y+1/z=1/2(2)1/z+1/x=1/3(3)[(1)+(2)+(3)]
【x+y】分之xy=-2,xy分之【x+y】=-1/21/x+1/y=-1/2(1)【y+z】分之yz=3分之4,yz分之【y+z】=3/41/y+1/z=3/4(2)【z+x】分之zx=-3分之4,
答案是11分之14. 求采纳
假设x=m+nc=m+(-1+√17)n/2y=p+qc=p+(-1+√17)q/2那么xy=[m+(-1+√17)n/2][p+(-1+√17)q/2]=mp+(9-√17)nq/2+(-1+√17
1/x-1/y=3(y-x)/xy=3y-x=3xy分式的分母=x-2xy-y=(x-y)-2xy=-5xy分子=2x+3xy-2y=2(x-y)+3xy=-3xy所以原式=3/5
x分之3=y分之1x=3yy分之1=z分之2z=2yxy+yz+zx分之2x²-2y²+5z²=[2(3y)²-2y²+5(2y)²]/(3
令3/x=4/y=6/z=1/k则x=3ky=4kz=6k(x²+y²+z²)/(xy+yz+xz)=(9k²+16k²+36k²)/(12
a^x=(ab)^z=a^z*b^za^(x-z)=b^zb=a^[(x-z)/z](1)b^y=(ab)^z=a^z*b^zb^(y-z)=a^zb=a^[z/(y-z)](2)(1)=(2)所以a
1/x+1/y=3(1)1/y+1/z=2(2)1/x+1/z=1(3)(1)+(2)+(3)2(1/x+1/y+1/z)=61/x+1/y+1/z=3(4)由(1)1/z=0题目有误,请核对,或者更