已知x y=7,xy=-8求x的平方 y的平方
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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
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