x y=1,y-z=-2
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(x+y+z)²=1,x²+2xy+y²+2(x+y)z+z²=1,x²+y²+z²+2(x+y)z+2xy=1xy+yz+xz=
1=xy/(x+y)两边倒数1/x+1/y=1同理1/y+1/z=1/21/z+1/x=1/3联合三个方程得1/x=5/121/y=7/121/z=-1/12即x=12/5y=12/7z=-12x+y
因为|x-y|>=0,根号(2y+z)>=0,z²-z+1/4=(z-1/2)²>=0所以要使式子的值为0,必须各项的值都为0所以x-y=0,2y+z=0,z-1/2=0解得z=1
x-3=y-2x-y=1y-2=z-1y-z=1x-3=z-1z-x=-2x^2+y^2+z^2-xy-yz-xz=x(x-y)+y(y-z)+z(z-x)=x+y-2zx-3=z-1y-2=z-12
xy+yz+xz={(x²+y²+z²+2xy+2xz+2yz)-(x²+y²+z²)}\2={(x+y+z)²-(x²
对方程e^(-xy)+2z-e^z=2两边微分,有:e^(-xy)*d(-xy)+2*dz-e^z*dz=0-e^(-xy)*(x*dy+y*dx)+2*dz-e^z*dz=0移项,得:(e^z-2)
∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
(xy+yz+xz)²=x²y²+x²z²+y²z²+2xyz²+2x²yz+2xy²z=1=x
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----
y=-12;一共是三个方程,因为xy/(x+y)=3推出(x+y)/(xy)=1/3-------方程1;同理:(y+z)/(yz)=1/2-------方程2;(x+z)/(xz)=1-------
(x+y+z)²=1²x²+y²+z²+2xy+2yz+2xz=1x²+y²+z²+2(xy+yz+xz)=1x&sup
题目是这样吧1=xy/(x+y),2=yz/(y+z),3=xz/(x+z)倒数法,写成每个式子的倒数;1=1/x+1/y,(1)1/2=1/y+1/z,(2)1/3=1/x+1/z(3)三式相加,得
仔细观察题目后会发现,等式的右边是不为零的整数,这样无法判断XYZ的值所以用加减消元法,将这几个等式变形,变为右边=0的另外几个等式,然后再因式分解.这样为从新列出关XYZ的三元一次方程组吧.然后解出
1/Y+1/X=1(1)1/Z+1/Y=2(2)1/X+1/Z=3(3)(1)+(2)+(3):1/X+1/Y+1/Z=3(4)(4)-(1):1/Z=2Z=1/2(4)-(2):1/X=1X=1题目
xy/(x+y)=1=>xy=x+y=>1/x+1/y=1--式一yz/(y+z)=2=>yz=2y+2z=>1/y+1/z=1/2--式二xz/(x+z)=3=>xz=3x+3z=>1/x+1/z=
xy/(x+y)=1=>(x+y)/(xy)=1=>1/x+1/y=1同理1/y+1/z=1/2;1/z+1/x=1/3联立求得1/x=5/121/y=7/121/z=-1/12所以(1/x)(1/y
很简单,当未知数在指数位置时用a^x=Ina*a^x但当未知数在指数和底数位置时,不能用a^x=Ina*a^x所以你一开始就错了z=(1+xy)^ylnz=yln(1+xy)(1/z)(dz/dy)=
|x-3|+|y+z|+|2z+1|=0则|x-3|=0x=3|y+z|=0y=-z=1/2|2z+1|=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4
(x+y+z)²=x²+y²+z²+2xy+2yz+2xz所以可得:xy+yz+xz=[(x+y+z)²-(x²+y²+z
把x=y+根号2代入得2y^2+2根号2y+2根号2*z^2+1=02[y+(根号2)/2]^2+2根号2*Z^2=0∴y+(根号2)/2=02根号2*z^2=0∴y=-(根号2)/2z=0x=(根号