(1){x 2=y 3=z 4,x 2y 3z=40

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

分3种情况A.Ky1>y3B.K>0时,y3>y1>y2C.K=0时,y1=y2=y3=0;绘函数图象可以直观得出结论

C.根据图像在二、四象限可得到

y3>y1>y2画出坐标系中的图像,就能看出

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

（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

x2-x1=(b-a)/3y3-y1=2(y3-y2)=2(b-a)/4所以(y3-y1)/(x2-x1)=3/2

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

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

x1/y1=x2/y2=x3/y3=1/2y1=2x1,y2=2x2,y3=2x3（x1+x2-x3)/（y1+y2-y3）=﹙x1+x2-x3)／[2﹙x1+x2-x3)]＝½

设x1+y1=x2+y2=x3+y3=x4+y4=X5+y5=A,y1=A-x1,y2=A-x2,...,y5=A-x5.yˉ=(y1+y2+...+y5)/5=(A-x1+A-x2+...+A-x5

先两组基之间的过渡矩阵P(y1,y2,y3)=(x1,x2,x3)P则T(x1,x2,x3)=(y1,y2,y3)=(x1,x2,x3)PT(y1,y2,y3)=T(x1,x2,x3)P=(y1,y2

∵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．

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．

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．

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

解 由x2＝y3＝z4，设x=2k，y=3k，z=4k，（1）x−2yz＝2k−6k4k＝−1，（2）x+3＝z−y化为2k+3＝k，∴2k+3=k2，即k2-2k-3=0，∴k=3或k=-