已知x y=-7,xy=12,求y根号x y x根号y x的值-搜答案-智慧作业帮

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^2+3xy+2y^2)/(x^2y+2xy^2)=(x+2y)(x+y)/[xy(x+2y)]=(x+y)/(xy)将x+y=7,xy=12代入(x+y)/(xy)=7/12

(x²+xy-12)²+(xy-2y²-1)²=0由于平方数都大于或等于0,所以上式成立的前提是：(x²+xy-12)²=0,即：x&sup

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

(4xy+12y)+(7x-(3xy+4Y-x))=4xy+12y+(7x-3xy-4y+x)=4xy+12y+7x-3xy-4y+x=(4-3)xy+(12-4)y+(7+1)x=xy+8y+8x当

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

∵x2+xy=-3，xy+y2=7，∴（x2+xy）+（xy+y2）=-3+7=4，即x2+2xy+y2=4．

答案是3/2你是不是把分母打错了教你个方法因为上下是齐次的直接令x=3y=2带入就行

x^2y+xy^2=xy（x+y）=7x6=42再问：��(x+1)^2=13��x^2+2x+3=再答：x^2+2x+3=（x+1）^2+2=13+2=15

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

最简单的方法就是用特殊值,令x=-2,y=1然后代入所求表达式,求出其值.为-8；另外此内题的另一个解法是变化所求表达式,使它变成关于xy^2的表达式.

x²-7xy+12y²=(x-4y)(x-3y)=0x=4y或x=3yx=4y时(x²-xy+y²)/2xy=(16-4+1)y²/8y²=

3xy=2x+3y+5(3y-2)(x-1)=7所以3y-2=7x-1=1得y=3x=2xy=6或3y-2=1x-1=7得y=1x=6xy=6所以xy=6

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x^2-7xy+12y=0,且x.y不等于0(x-3y)(x-4y)=0x=3y或x=4y(x^2-2xy+y^2)/(2xy)=(x-y)^2/(2xy)x=3y时原式=4y^2/(6y^2)=2/

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

原式=-xy²（x²y^4-xy²-1）∵xy²=-2原式=2（（-2)²-（-2）-1）=10