x=5 y=5 z=5 x%=y z
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![x=5 y=5 z=5 x%=y z](/uploads/image/f/892616-32-6.jpg?t=x%3D5+y%3D5+z%3D5+x%25%3Dy+z)
xy/x+y=4/5,则x+y/xy=5/4,则1/y+1/x=5/4①yz/y+z=20/9,则y+z/yz=9/20,则1/z+1/y=9/20②xz/z+x=5/6,则z+x/xz=6/5,则1
(x+y+z)^2=25x^2+y^2+z^2+2*(x+y+z)=25z^2=23-(x^2+Y^2)0
xy/x+y=4/5,则x+y/xy=5/4,则1/y+1/x=5/4①yz/y+z=20/9,则y+z/yz=9/20,则1/z+1/y=9/20②xz/z+x=5/6,则z+x/xz=6/5,则1
题目应为:xy/(x+y)=6/5yz/(y+z)=12/7xz/(x+z)=4/3求x和y和z运用倒数变形可解因为1/y+1/x=5/6,1/z+1/y=7/12,1/z+1/x=3/4三式相加得1
若3/x=2/y=5/z,则x/3=y/2=z/5,设x/3=y/2=z/5=K,则x=3K,y=2K,z=5K将其代入上式,得6k2+10k2+15k231k2-------------------
xy/(x+y)=51/x+1/y=1/5yz/(y+z)=7/21/y+1/z=2/7zx/(z+x)=41/x+1/z=1/4(xy+yz+zx)分之xyz=1/(1/x+1/y+1/z)=280
x-y=5,z-y=10相减z-x=5x²+y²+z²-xy-yz-xz=(2x²+2y²+2z²-2xy-2yz-2xz)/2=[(x&s
X=1,Y=2,Z=3其实很简单!
2x-y-5z=0,x-2y+2z=0,3x-12z=0;x=4z;y=3z;x²+y²+z²/xy+yz+zx=(16z²+9z²+z²)
∵x+y+z=5∴x=5-y-z∵xy+yz+xz=3∴y^2+(z-5)y+(z^2-5z+3)=0又∵y,z是实数,∴△=(z-5)^2-4(z^2-5z+3)=(z+1)(-3z+13)≥0∴-
xy\X+Y=12\71/y+1/x=7/12(1)YZ\Y+Z=6\51/z+1/y=5/6(2)XZ\X+Z=4\31/z+1/x=3/4(3)由(1)-(2)得1/x-1/z=-1/4(4)由(
x=1,y=2,z=3
将x=5-y-z代入xy+yz+zx=3,整理成关于y的一元二次方程y²+(z-5)y+z²-5z+3=0由于y为实数,所以△≥0.即(z-5)²-4(z²-5
由X+Y+Z=5得Y=5-X-Z将此代入XY+YZ+ZX=3得X(2-X-Z)+(5-X-Z)Z+ZX=3整理得X^2+(Z-5)X+(Z^2-5Z+3)=0因为X是实数,那么关于X的一元二次方程的判
很久没做过,不知道我做的对不对,参考一下吧x+y+z=5,xy+xz+yz=3.但是(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)所以x^2+y^2+z^2=19.x^2+y^2=
(x+y+z)^2=4x^2+y^2+z^2+2xy+2xz+2yz=4x^2+y^2+z^2+2(-5)=4x^2+y^2+z^2=14
x+y+z=5,xy+yz+zx=9所以(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25所以x^2+y^2+z^2=25-2×9=25-18=7
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
xy+yz+zx=93中的y,z全用x代替可以得到2x^2/3+10x^2/9+5x^2/3=93∴x^2=27同理y^2=12z^2=75∴9x*x+12y*y+2z*z=9*27+12*12+2*
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25x^2+y^2+z^2=25-14=11