已知x≠y,且x^2-x=3,y^2-y=3,求代数式x^2-xy y^2的值-搜答案-智慧作业帮

x=3yx+y=12所以x=9,y=3x-y=9-3=6

x*x+√3y=√5,y*y+√3x=√5x*x+√3y=y*y+√3xx*x+√3y-y*y-√3x=0(x-y)(x+y)-√3(x-y)=0(x-y)(x+y-√3)=0x≠y,x-y≠0x+y

令x＝y＝0得2f(0)=2f^2(0),于是f(0)=0.（因为f(0)不为1）.再令x＝0得f(y)+f(－y)=2f(0)f(y)=0,因此f(－y)＝－f(y),f是奇函数.显然有F(－x)=

由已知：xy+x+y=17,xy(x+y)=66,可知xy和x+y是方程t2-17t+66=0的两个实数根,得：t1=6,t2=11.即xy=6,x+y=11,或xy=11,x+y=6.当xy=6,x

7

LZ你确定题目是这样的?xyz≠0可以推出XYZ中任意一个数不为02x+3y=3x-8y可以推出x=11y这样就可以直接代入上式了(x+y)/(x-y)=6/5我想不到z有什么用或者我欠考虑了……

2x+3y=3x-8yx=11y12y/10y=6/5确定题没给错?

X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1

|x|=根号3-根号2|y|=根号2当x,y同时为正时x=根号3-根号2y=根号2x+y=根号3题意x+y≠根号3所以不可能同时为正.当x正,y为负时x=根号3-根号2y=-根号2x+y=根号3-2根

y=3-x带入2x+y=zx+3=z带入中间的式子3x+12-4x=2x+6+6x=0y=3z=3x=0Z=3

因为(x-y)/(x+y)=3,则(x＋y)/(x－y)＝1/3则5(x-y)(x+y)-(x+y)/2(x-y)＝5*3-1/(3*2)=15-1/6=89/6

c∵x+y≠0∴x≠-y∵3x-6y/x+y=0∴3x-6y=0∴x=2y∴x=2y且x≠-y

∵2x－6≥0,x≥33－x≥0.x≤3∴x＝3,y＝1x·√2x÷√﹙x/y﹚＝x·√2x·√﹙y/x﹚＝x·√2y＝3×√2×1＝3√2

根据题意得：x-2≥02-x≥0，解得：x=2，则y=3，故2x-y=1，则平方根是：±1．

(x+y)(3x-4y)=0所以y=-x或y=3x/4y=-x原式=(x+2x)/(x-2x)=-3y=3x/4原式=(x-3x/2)/(x+3x/2)=-1/5

令x除以2=y除以3=z除以4=k则x=2k,y=3k,z=4kx+y+z除以x-2y+3z=(2k+3k+4k)除以(2k-6k+12k)=9k除以8k=8分之9

2x²-xy-3y²=0(x+y)(2x-3y)=0x=-y(舍去)或x=3y/2x/y=3/2x/y-x²/(x-y)=x/y-x²/(x²-2xy

x^2=12x=±2√3y^2=3y=±√3|x-y|=y-x说明y-x≥0即y≥x所以x不可能等于2√3,即x=-2√3若y=-√3,x+y=-3√3若y=√3,x+y=-√3

2X^2+3xy-2y^2=0(2x-y)(x+2y)=0即：2x-y=0或,x+2y=0y=2x或x=-2y当y=2x时：x+y/x-y=3x/(-x)=-3当x=-2y时：x+y/x-y=-y/(

2x²-xy-3y²=0(x+y)(2x-3y)=0x=-y,2x=3yx≠-y所以y=(2/3)x原式=(x²+2/3*x²)/(2/3*x²-4/