x z-3=0 2x-y 2z=2 x-y-z=-3
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![x z-3=0 2x-y 2z=2 x-y-z=-3](/uploads/image/f/888919-7-9.jpg?t=x+z-3%EF%BC%9D0+2x-y+2z%EF%BC%9D2+x-y-z%3D-3)
=[(X+Z)+(X-Y)]/[X(X-Y)+Z(X-Y)]-[(X+Y)+(X+Z)]/[X(X+Y)+Z(X+Y)]=[(X+Z)+(X-Y)]/[(X+Z)(X-Y)]-[(X+Y)+(X+Z)
xy/(x+y)=1,取倒数(x+y)/xy=1x/xy+y/xy=11/y+1/x=1.1yz/(y+z)=2,取倒数(y+z)/yz=1/2y/yz+z/yz=1/21/z+1/y=1/2.2xz
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
x²+y²+z²=xy+yz+xz两边各乘以2得到2x²+2y²+2z²=2xy+2yz+2xzx²-2xy+y²+x&
X^2+Y^2+Z^2=XY+YZ+XZ则有2X^2+2Y^2+2Z^2-2XY-2YZ-2XZ=0==>(X-Y)^2+(Y-Z)^2+(Z-X)^2=0必然X-Y=0,Y-Z=0,Z-X=0==>
结果等于:1原式=x/(xy+x+xyz)+y/(yz+y+xyz)+z/(xz+z+xyz)=1/(y+1+yz)+1/(z+1+xz)+1/(x+1+xy)=xyz/(y+xyz+yz)+1/(z
x²-xy+xz-yz=x(x-y)+z(x-y)=(x+z)(x-y)若仍有疑问,欢迎追问!
应该是设X/2=Y/1=Z/3=K则X=2KY=KZ=3K则有xy+xz+yz=992K^2+6K^2+3K^2=99==>K^2=9所以4x^2-2xz+3yz-9y^2=2X(2X-Z)+3Y(Z
thedragon53的错了,(1)-(2)得2xz-yz=4,而不是2xz+yz=4正确的做法:xy=xz+3.①,yz=xy+xz-7.②(x,y,z均为正整数)由①得到y=z+3/x,由于x,y
令x/3=y/2=z/5=k则x=3ky=2kz=5k∴(xy+yz+zx)/(x²+y²+z²)=(6+10+15)k²/(9+4+25)l²=31
可以联立条件解得y=3z,x=4z,然后代入原式使用最猥琐的方法求解即可,由於是齐次式所以元一定会消掉的,结果我口算的是19分之26,不知道对不对
因为几个非负数的和为0时必有每个非负数都为0.而一个数的算术平方根是非负数,所以√(2x+3)+√(4y-6x)+√(x+y+z)=0时,有√(2x+3)=0且√(4y-6x)=0且√(x+y+z)=
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
原式=[(x--y)+(x--z)]/(x--y)(x--z)+[(y--x)+(y--z)]/(y--x)(y--z)+[(z--x)+(z--y)]/(z--x)(z--y)=1/(x--z)+1
(1)原式=-6x3y3z+4x2y3z;(2)原式=4a4b2-4a2b4-4a4b4÷4b2+4a2b4=3a4b2;(3)原式=1232-(123+1)×(123-1)=1232-(1232-1
答案是:(2*X)/((X-Z)*(X+Z))再问:解题过程给我写下1再答:=(2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)
令2/x=3/y=7/z=k∴x=2/ky=3/kz=7/k∴(xy+xz+yz)/(x^2+y^2+z^2)=(2/k*3/k+2/k*7/k+3/k*7/k)/(4/k²+9/k
若x与z互为相倒数,则xz=1,|y|=7,y=±7,则xz+y=1±7=8or-6.
(x+2y-z)^2+(z-x)^2=0所以x+2y-z=0,z-x=0x=z所以2y=0,y=0代入xz^2+yz-5√(xz^2+yz+9)+3=0x^3-5√(x^3+9)+3=0(x^3+9)