{x-y-5z=4 2x y-3z=10 3x y z=8
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x+y=5x=5-yz^2=xy+y-9z^2=(5-y)y+y-9z^2=-y^2+6y-9z^2=-(y-3)^2z^2+(y-3)^2=0所以,z=0,y-3=0z=0,y=3x=5-y=5-3
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-------------------
x-y=5x=5+yz^2=-xy-y-9=-(5+y)y-y-9=-y^2-6y-9=-(y+3)^2所以,z=0,y+3=0z=0,y=-3x=5+y=5-3=2x-2y+3z=2-2*(-3)+
X=1,Y=2,Z=3其实很简单!
∵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+y=5,z2=xy+y-9,∴x=5-y,代入z2=xy+y-9得:z2=(5-y)y+y-9,z2+(y-3)2=0,z=0,y-3=0,∴y=3,x=5-3=2,x+2y+3z=2+2×3+
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=
由4x-5y+2z=0,(1)x+4y-3z=0,(2)将2式乘以4减去1式,可以得出,21y=14z,即z=1.5y代回1式可得,4x-5y+3y=0,即4x=2y,x=0.5y分别代入(x
x=5-yz2=(5-y)y+y-9=6y-y2-9=-(9-6y+y2)=-(y-3)2由题意,只有当该项为0时等式成立得y=3那么z=0x=2即原式=2+6+0=8
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*
平方和绝对值都大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个式子都等于0所以x-3y+z=0(1)5x-4y+z=0(2)(1)-(1)4x-y=0y=4x(2)-(1)*5
时间太长不是太会做不过希望对你有帮助9z²=z²+25y²-10yz9z²-25y²=z²-10yz所以x²-25y²+
解方程组:{2x-3y-z=0.(1){x+3y-14z=0.(2)(1)+(2)得:3x-15z=0即:x=5z,代入(1)式得y=3z所以:(4x²-5xy+z²)/(xy+y
x+3y+7z=02x+5y+11z=0x=2z,y=-3zz=0,x=y=0分式无意义(x^2+y^2+z^2)/(xy+2yz+3xz),z≠0=[(2z)^2+(-3z)^2+z^2]/[(-2