作业帮分别求Y=2X-1,Z=X2的分布律
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/10 15:35:32
2x+3Y=4z6x+9y=12z(1)3x+4y=-5z6x+8y=-10z(2)(1)-(2)得y=22z带入（1）得x=-31z（2）x+y+z=-8zx-y+z=-52z答案为-1/13再问：
利用柯西不等式∵（x^2+y^2+z^2）（2^2+3^2+4^2）≥（2x+3y+4z）^2∴x^2+y^2+z^2≥（2x+3y+4z）^2/（2^2+3^2+4^2）=100/29当x/2=y/
因为x:y:z=3：4：5所以设x=3k,y=4k,z=5k(k≠0)(1)z/(x+y)=5k/(3k+4k)=5k/7k=5/7(2)x+y+z=63k+4k+5k=612k=6k=1/2x=3k
∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
1.x^2+y^2+z^2=(x+y+z)^2-2*(xy+yz+xz)=82.配方(x-1)^2+(y+2)^2+(z-3)^2=0x=1y=-2z=3x+y+z=2再问：兄弟我敬佩你，如果你能帮我
1.(x-1)^2+(y+2)^2+(z-3)^2=0则x＝1,y＝－2,z＝3x+y+z＝22.(3a－2b)(a＋b)＝0则a＝－b或a＝2/3×b则a/b-b/a-(a^2+b^2)/ab＝(a
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/>x^2+4y^2+z^2-2x+4y-6z+11=0(x²－2x＋1)＋(4y²＋4y＋1)＋(z²－6z＋9)=0(x－1)²＋(2y＋1)²＋
x：y：z=1：2：3,x=k,y=2k,z=3kx+y+z=k+2k+3k=6k=12k=2x=2,y=4,z=6
∵x-z=（x-y）+（y-z）=2+2=4∴x2-z2=（x+z）（x-z）=14×4=56．
x/(y+z)+y/(z+x)+z/(x+y)=1所以x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+
等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+
x+y大于等于2倍根号下xy同理x+z大于等于2倍根号下xzz+y大于等于2倍根号下zy所以(x+y)(y+z)(z+x)大于等于8xyz当取到8xyz时分数值最大为1/8此时x=1/3y=1/3z=
∵x=-2，x+y+z=-2.8∴y+z=-0.8原式=（z+y）（-x2-3.2x）=（-0.8）（-4+6.4）=-0.8×2.4=-1.92
x+y+z=1xy+yz+zx=21*2=(x+y+z)(xy+yz+zx)=x(xy+yz+zx)+y(xy+yz+zx)+z(xy+yz+zx)=x²y+xyz+zx²+xy&
①令x/2=y/4=z/7=k所以x=2k,y=4k,z=7k所以x：y：z=2:4:7②2x+y+3z=58即4k+4k+21k=5829k=58k=2x=2k=4y=4k=8z=7k=14③(x+
换元.可设x=a+b,y=a-b.则z=2(a²+b²)-(a²-b²)-2(a+b)+(a-b)=a²-a+3b²-3b=[a-(1/2)
(x+y+z)(x+y+z)=a2=a2/2+2xy+2xz+2yz,有a2/2=2x(y+z)+2yz=2x(a-x)+2yz,则有a2/2-2ax+2x2=2yz(由于2yz