已知实数x,y,z满足三分之x=一分之y=二分之z
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x+2y-z=6①x-y+2z=3②,①×2+②,得x+y=5,则y=5-x③,①+2×②,得x+z=4,则z=4-x④,把③④代入x2+y2+z2得,x2+(5-x)2+(4-x)2=3x2-18x
用x来表示y和z解方程组y-z=-x-y+2z=-3x两式相加得z=-4x把z=-4x代入y-z=-x中,得y=-5x所以x:y:z=x:(-5x):(-4x)=1:(-5):(-4)或x:y:z=-
实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.
设x/2=y/3=z/4=k∴x=2k,y=3k,z=4k∵2x-3y+4z=22∴4k-9k+16k=22∴k=2∴x=4,y=6,z=8∴x+y-z=4+6-8=2
把x=6-y带入z^2-4z+4=xy-9中,得(y-3)^2+(z-2)^2=0,故y-3=0,z-2=0,所以y=3,z=2,x=3.
1.x/2=y/3=z/41)x+y+z=12)1)式变形,得y=3x/2z=2x代入2),dex+3x/2+2x=1x=2/9y=1/3z=4/92.两个非负数互为相反数,那么这两个数都为02x+y
设二分之x=三分之y=二分之一分之z=k则x=2ky=3kz=2kx+3y-z分之2x-y+z=9k分之3k=1/3再问:第三个是二分之一分之z,那么z不应该=2k吧?最后答案是7,我算不出结果再答:
已知二分之x=三分之y=四分之z∴y=3x/2;z=2x;3x-2y+z分之2x+y-z=(2x+y-z)/(3x-2y+z)=(2x+3x/2-2x)/(3x-3x+2x)=3x/2/2x=3/4
x/2=y/3则3x=2yy/3=z/4则z=4y/3所以2y-4y+20y/3=14y=3所以z=4x=2
x+2y-z=6,.(1)x-y+2z=3.(2)(1)-(2)y-z=1,y=1+z(1)+2(2)x+z=4,x=4-zx^2+y^2+z^2=(4-z)^2+(1+z)^2+z^2=3z^2-6
题目不明确,但大体是是这样吧3
XY/X+Y=-2,-->(x+y)/(xy)=-1/2,-->1/x+1/y=-1/2YZ/Y+Z=4/3,-->(y+z)/(yz)=3/4,-->1/y+1/z=3/4ZX/Z+X=-4/3,-
因为x/y+z+y/z+x+z/x+y=1所以x/y+z=1-y/z+x-z/x+y,两边同乘以x得x^2/y+z=x-xy/z+x-xz/x+y同理y^2/x+z=y-xy/z+y-yz/x+y,z
x+y-z/z=y+z-x/x=z+x-y/y,应用等比定理,得(x+y-z+y+z-x+z+x-y)/(x+y+z)=(x+y-z)/z,所以(x+y+z)/(x+y+z)=(x+y-z)/z,即1
2x-3y-z=0..(1)x-2y+z=0...(2)(1)+(2):3x-5y=03x=5yy=3/5x将y=3/5x代入(1)z=2x-3y=2x-3*3/5x=x/5x:y:z=x:3/5x:
等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+
z^2>=0-z^2=0所以只会存在z^2=0,也就是z=0x²+4y²+根号-z²=2x+4y-2就可以简化成x^2+4y^2=2x+4y-2移项就得到x^2-2x+1
∵正实数x,y,z满足2x(x+1y+1z)=yz,∴x2+x(1y+1z)=12yz,∴(x+1y)(x+1z)=x2+x((1y+1z)+1yz=12yz+1yz≥212=2.当且仅当yz=2,取
令a=x-yb=y-z则z-x=-(a+b)所以原条件即为(a+b)^2-4ab=0(a-b)^2=0所以a=b所以x-y=y-z这说明x,y,z是等差数列
(z-x)²-4(x-y)(y-z)=0.z²+x²-2xz-4(xy-xz-y²+yz)=0z²+x²+2xz-4xy+4y²-