圆周x^2 y^2 z^2=0,求x^2 xds
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由x+y-z=0,2x-y+2z=0可得:z=-3xy=-4x则3x-2y+5z/5x-3y+2z=3x+8x-15x/5x+12x-6x=-4x/11x=-4/11
x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1
因为:X+Y+Z=0得:Z+Y=-X------(1)X+Y=-Z------------(2)Z+Y=-X------------(3)X^3+X^2Z-XYZ+Y^2Z+Y^3=X^3+XZ(X+
x+y+z-6=02x+3y-z-12=02x-y-z=0组成方程组再解x=2y=3z=1
本题不一定要算出x、y,也能解出.(1)2x+2y-4z=0(2)2x+4y+5z=0(2)-(1):2y+9z=0即2y=-9z(1)÷2可得:x+y=2z故x+y+z=3z则x-y+z=(x+y+
因为x+2y-z=0,7x-y-z=0两式相减,得:6x-3y=0,所以y=2x代入x+2y-z=0中,得:x+4x-z=0,那么z=5x那么(x+y+z)÷(2x-y-z)=(x+2x+5x)÷(2
1.z²-z+1/4=(z-1/2)².绝对值、根号、平方数都是非负的,而相加为0.所以都为0.即x=y,2y=z,z=1/2.所以x=y=1/4,z=1/2.2.2002x200
0和-2再答:0和-2再问:说清是X等于几和z等干几
/>x+y-z=6①x-3y+2z=0②3x+2y-z=0③由③-①得:2x+y=-6④由②+③×2得7x+y=0⑤由⑤-④得:5x=6x=6/5把x=6/5代入⑤得:42/5+y=0y=-42/5把
由2x+3y-3z=0得:z-y=2x/3(2x+y-z)/(2x-y+z)=(2x-(z-y))/(2x+(z-y))将z-y=2x/3代入上式得:(2x+y-z)/(2x-y+z)=(2x-(2x
4x-3y+z=0(1)x+2y-8z=0(2)(1)-(2)×4得-11y+33z=0∴y=3z把y=3z代入(2)得x=2z把x=2z,y=3z代入x+y-z/x-y+2z得原式=(2z+3z-z
x²+4x+4+y²+4y+4+z²+4z+4=-1+4+4+4(x+2)²+(y+2)²+(z+2)²=11[(2-(-x))²
稍等.再问:……我一直等着再答:这个题目不太对,应该是求X+Y+Z的最小值吧,再问:你的想法是什么?再答:因为x+y+z的值有无穷个答案。。。再问:你是怎么推算的?再问:我是想问这个再答:这很简单啊,
x²+4x+4+y²+4y+4+z²+4z+4=-1+4+4+4(x+2)²+(y+2)²+(z+2)²=11[(2-(-x))²
X+Y-Z=6.aY+Z-X=2.bZ+X-Y=0.ca,b,c三式相加X+Y+Z=8.dd式-a式2Z=2Z=1d式-b式2X=6X=3d式-C式2Y=8Y=4
括号是什么意思?只有一半
两式相加,得3x-z=0可得z/x=5将z=5x代入1式13x-y=0得y/x=13所以x:y:z=1:13:5
x*x+y*y+2z*z-2x+4y+4z+7=0(x*x-2x+1)+(y*y+4y+4)+2(z*z+2z+1)=0(x-1)^2+(y+2)^2+2(z+1)^2=0x=1,y=-2,z=-1x
令a=x/2=y/3=z/4则x=2a,y=3a,z=4a所以原式=(2a+3a+4a)/(2a+3a-4a)=9a/a=9
x+y-7z=0①x-2y+5z=0②①-②得:3y-12z=0,即y=4z,2①+②得:3x-9z=0,即x=3z所以x+2y-z/y-2x+12z=11/10,再来题难点的.