x-2y+z 9=2x+y+3z 10=3x+2y﹣4z 3=1
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1.x+y+z2.-x-3y-1
(y-z)^2+(z-x)^2+(x-y)^2=(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2[(y-z)^2-(y+z-2x)^2]+[(z-x)^2-(x+z-2y)^2]+[(
∑是循环和例如∑a=a+b+c∑a^2=a^2+b^2+c^2∑(z-y)(x-y)/(x+y-2z)(y+z-2x)=∑(z-y)(x-y)(x+z-2y)/(x+y-2z)(y+z-2x)(x+z
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=4x-2y-z-5x[6y-(x+y)]+x-(3y-10z)=4x-2y-z-30xy+5x²+5xy+x-3
1.x=1,y=2,z=-12.x=-1,y=-3,z=23.a=-5/2,b=7/2,c=2其他的我也不说了,慢慢想吧~
有这样的公式:a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3(x+y
正整数?取对数即证:2xlnx+2ylny+2zlnz>(y+z)lnx+(x+z)lny+(x+y)lnzx>y>z,lnx>lny>lnz由排序不等式得xlnx+ylny+zlnz>ylnx+zl
x+y+z-6=02x+3y-z-12=02x-y-z=0组成方程组再解x=2y=3z=1
【解】视z为常数,由已知两方程,可解得x=3zy=2z将其代入待求值式中,得3x*x+2y*y+5z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+5z^2]/[5(3z)^2+(
设a=x-y,b=y-z,-a-b=z-x(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方b^2+a^2+(-a-b)^2=(-a-b-
证:(y+z-2x)³+(z+x-2y)³+(x+y-2z)³=[(y+z-2x)³+(z+x-2y)³]+(x+y-2z)³/前面两项运用
1、隐函数对x求导得1+az/ax+yz+xy*az/ax=0,故az/ax=-(1+yz)/(1+xy);F对x求导得aF/ax=e^x*y*z^2+e^x*y*2z*az/ax;当x=0,y=1时
若a+b+c=0,可证a^3+b^3+c^3-3abc=a^3+b^3+(-a-b)^3-3ab(-a-b)=a^3+b^3+3ab(a+b)-(a+b)^3=(a+b)^3-(a+b)^3令a=x+
根号x-3+|y-2|+z^2=2z-1根号x-3+|y-2|+(z^2-2z+1)=0根号x-3+|y-2|+(z-1)^2=0由于数值开根号,绝对值和平方数均为大于等于0的数则上式要成立只有X-3
x+2y=32y=3zx-y=-1x+2y=3①2y=3z②x-y=-1③①-③得3y=4,得y=4/3代入③,得x=y-1=1/3代入②,得z=2/3y=8/9x+y+z=36x-y=12x+z-y
=x²(y-z)+y²(z-x)+z²(x-z+z-y)=(y-z)(x²-z²)+(z-x)(y²-z²)=(y-z)(x-z)
3元一次方程,好像是初一的问题哦.根据前面两个等式可以得出x=3zy=z(平方)/32x+3y+4z=2*(3z)+3*(z方/3)+4z现在变成了一元二次方程,你应该会解吧.
设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是:(x+3y-z)/(x-3y+z)?若是,如下:(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3
根据公式(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac公式展开:得到(x^2+y^2+z^2=2xy-2yz-2xz)-(x^2+y^2+z^2-2xy-2yz+2xz)合并同类项