已知x y=7,xy=-8求x的平方 y的平方-搜答案-神马作文网

由X+Y=3XY可得X+Y-XY=2XY,用X+Y-XY替换分母中的2XY,所以(2X+2Y-XY)/(X+Y+2XY)=(2X+2Y-XY)/(2X+2Y-XY)=1

题目写错应该是求（6xy+7y)+[8x-(5xy-y+6x)]（6xy+7y)+[8x-(5xy-y+6x)]=6xy+7y+8x-5xy+y-6x=xy+2x+8y=xy+2(x+4y)=5+2×

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

令F（XY）=1/XY+XY,当XY=1的时候,F（XY)=2,最小.（可由函数图形象得出）.XY趋于正无穷大的时候F(XY)趋于正无穷大,XY无限趋于零的时候F(XY)趋于正无穷大.所以XY越接近1

x-y=2xy3x-5xy-3y/x+xy-y=[3(x-y)-5xy]/[(x-y)+xy]=(6xy-5xy)/(2xy+xy)=1/31/x+1/y=4y+x=4xyx-5xy+y/2x+xy+

xy/x+y=3xy=3(x+y)3x-5xy+3y/-x+3xy-y=(3x+3y)-5*3(x+y)/[-x-y+3*3(x+y)]=-12(x+y)/8(x+y)=-3/2望采纳,谢谢!

-xy（x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10

原式=6x+4xy+2xy+3x+2y-4xy+7y=9x+9y+2xy=9(x+y)+2xy=9×4+2×2=36+4=40

5xy+4x+7y+6x-3xy-4xy+3y=5xy-3xy-4xy+4x+6x+7y+3y=（5-3-4）xy+（4+6）x+（7+3）y=-2xy+10x+10y=-2xy+10（x+y）因为x

解题思路：本题是求整式的值，注意一定要先化简，再求值。化简就是先去括号，在合并同类项。解题过程：解：（6xy+7y）+[8x-（5xy-y+6x）]=（6xy+7y）+[8x-5xy+y-6x]=6x

原式=6xy+7y+8x-5xy+y-6x=xy+2x+8y=xy+2(x+4y)=···再问：还有些题哈……

原式=6xy+7y+8x-5xy+y-6x=xy+8y+2x=xy+2(x+4y)=5+2×(-1)=3

6xy+7y）+【8x-（5xy-y+6x）】=xy+8y+2x=xy+2(4y+x)=5-2=3

（6xy+7y)+[8x-(5xy-y+6x)]=6xy+7y+（8x-5xy+y-6x）=6xy+7y+8x-5xy+y-6x=（6xy-5xy）+（7y+y）+（8x-6x）=xy+8y+2x=x

先把代数式（5XY+4X+7Y)+(6X-3XY)-(4XY-3Y)化简（5XY+4X+7Y)+(6X-3XY)-(4XY-3Y)=5XY+4X+7Y+6X-3XY-4XY+3Y=-2XY+10X+1

16524

那9是负的吗～如果不是等于负1,如果是负九等于十七～用第一个等式减第二个等式

3(x^2-xy)-xy+y^2=3(x^2-xy)-(xy-y^2)=3*5-(-3)=15+3=18

x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式＝9y²-3y²+y²/6y²=7/6x=4y时原式

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60