2|x-y| √(2y z) z²-z ¼=0
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=[(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
f(x,y,z)=yz+xz使得,y^2+z^2=1,yz=3令F(x,y,z)=yz+xz+a(y²+z²-1)+b(yz-3)Fx=z=0Fy=z+2ay+bz=0Fz=y+x
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz>3(xy+yz+zx)所以只要求证x^2+y^2+z^2>xy+yz+zx2(x^2+y^2+z^2)>2(xy+yz+zx)(x^
=(x+y+z)^2+yz(y+z+x)=(x+y+z)(x+y+z+yz)
道理很简单:因为yz中的z也是y的函数,不可以只对y求偏导,必须对(yz)用积的求导方法求导,自然就是yz+yαz/α/x了.再问:那为什么这道题中是yαz/αx(e^z-xy)而不是[yαz/αx+
解x²-y²-z²+2yz=x²-(y²+z²-2yz)=x²-(y-z)²=(x+y-z)(x-y+z)
由题式可以看出当x=y或y=z或x=z时式子为0所以肯定有因式(x-y)(y-z)(z-x)展开后x最高项为-x^2y与x^2z而原式中x最高次项为x^3y和-x^3z所以还差x的1次项因式,所以实际
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
x-y=(1+√3)/2.(1)z-y=(1-√3)/2.(2)(1)-(2):(x-z)=√3.(3)(1)²+(2)²+(3)²得:(x-y)²+(z-y)
x+y+z=2√x+2√(y-1)+2√(z-2)[x-2√x+1]+[(y-1)-2√(y-1)+1]+[(z-2)+2√(z-2)+1]=0(√x-1)^2+[√(y-1)-1]^2+[√(z-2
√(x^2+xy+y^2)+√(y^2+yz+z^2)+√(z^2+zx+x^2)>=√(1/4*x^2+xy+y^2)+√(1/4*y^2+yz+z^2)+√(1/4*z^2+zx+x^2)=√(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)
1.=x^2-(y+z)^2=(x+y+z)(x-y-z)2.a^2-b^2+c^2-2ac=(a-c)^2-b^2=(a-c-b)(a-c+b)ac-b可知原式
解题思路:本题的关键是将三个方程两边取倒数,化简后分别将方程等号左边和右边相加,得到1/x+1/y+1/z的值,最后将要求的分式化简,把1/x+1/y+1/z的值带入即可。解题过程:
|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^2-yz)/[x^2-(y+z)x+yz]+(y^2-zx)/[y^2-(z+x)y+zx]+(z^2-xy)/[z^2-(x+y)z+xy]=(yz-x^2)/(x-y)(z-x)+(zx-y
①x:y:z因为xy:yz:zx=3:2:1所以xy:yz=3:2所以x:z=3:2同理yz:zx=2:1所以y:x=2:1=6:3所以x:y:z=3:6:2②x/yz:y/zx=x^2:y^2=(x