x 2=y 3=z 4,求2x^2-y^2 z^2 xy 2yz-3xz-搜答案-搜搜考试网

x=log2(y)则X1+2X2+3X3=log2(y1)+2log2(y2)+3log2(y3)=log2(y1)+log2(y2^2)+log2(y3^3)=log2(y1y2^2y3^3)=1所

∵y=-2/x在（负无穷,0）上是增函数∴当y1>y2>y3>0时,0>x1>x2>x3选C

x3+y3=100(x+y)(x^2-xy+y^2)=100因x+y=1所以x^2-xy+y^2=100(x+y)^2-3xy=1001-3xy=100xy=-33x^2+y^2=(x+y)^2-2x

因为:X3-Y3-Z3=3XYZ所以:X3+(-Y)3+(-Z)3-3X(-Y)(-Z)=0(X-Y-Z)(X2+Y2+Z2+XY+XZ-YZ)=0所以:1.X-Y-Z=02.X2+Y2+Z2+XY+

24

（x+y+z）²-（x²+y²+z²）=2（xy+yz+zx）=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&

（x+y+z）^2=[(x+y)+z]^2=(x^2+2xy+y^2)+z^2+2zx+2zy=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2(xy+xz+yz)=0x+y

∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2

设x2=y3=z4=k，则x=2k，y=3k，z=4k，∵2x-3y+4z=22，∴4k-9k+16k=22，∴k=2，∴x+y-z=2k+3k-4k=k=2．

y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得：y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以

设x+3y3＝2y+3z4＝2z+2x5=k，则有：x+3y＝3k2y+3z＝4k2z+2x＝5k，解得x＝32ky＝12kz＝k；因此x：y：z=3：1：2．

(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了

∵方程2x2-7xy+3y3=0有正整数解，∴△=49y2-24y3=y2（49-24y）≥0，且y＞0，解得，0＜y≤4924；∴y=1或y=2；①当y=1时，原方程化为2x2-7x+3=0，即（2

请想想直线方程通式y=kx+b三个点都在直线上,分别代入方程5=3k+b-------b=5-3k7=kx2+b-------kx2=7-5+3k=2+3k-----k=2----x2=4y3=-1k

7-5=2(x2-3),x2=4y3-5=2(-1-3),y3=-3

若x2＝y3＝z4

∵x2＝y3＝z4，∴6x=4y=3z，∵3x-2y+5z=-20，∴6x-4y+10z=-40，∴z=-4，∴x=-2，y=-3，∴x+3y-z=-2+3×（-3）-（-4）=-7；故答案为：-7．

已知x2＝y3＝z4

设x2＝y3＝z4=k，则x=2k，y=3k，z=4k，∴4x−3y+5z2x+3y=4×2k−3×3k+5×4k2×2k+3×3k=1913．故答案为：1913．

如果x2＝y3＝z4

根据题意，设x=2k，y=3k，z=4k∵x+y+z=18∴2k+3k+4k=18，解得k=2∴x=4，y=6，z=8∴x+y-z=2．

已知x2＝y3＝z4＝k

由题意得：x=2k，y=3k，z=4k，则原式=4k+3k−4k6k−6k+4k=34．

再问：能把第三题重新发一遍吗？再答：这三个题本质上式连在一起的再答：这道题应该是希望杯的试题