如果x-y=14,xy=2,求下列多项式的值-搜答案-神马作文网

答：x+y=-3,xy=-2x³y²+x²y³=(x+y)(xy)²=(-3)*(-2)²=-3*4=-12再问：确定么急的、再答：没有错误

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

x^2+xy+y=14,y^2+xy+x=28两式相加得x²+2xy+y²+(x+y)=42(x+y)²+(x+y)-42=0(x+y+7)(x+y-6)=0所以x+y+

(x^2+xy+2y^2)/(x^2-xy+y^2)=[(x/y)²+(x/y)+2]/[(x/y)²-(x/y)+1]分子分母同时除以y²=(4/9+2/3+2)/(4

x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7

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

上下都除以y^2

X/Y=4/3设X=4k,Y=3k,k≠0则(3X^2-5XY+2Y^2)/(2X^2+3XY-5Y^2)=(3*(4k)^2-5*(4k)*(3k)+2*(3k)^2)/(2*(4k)^2+3*(4

如果有理数xy满足丨x-1丨+（xy-2)²=0求x和y的值丨x-1丨+（xy-2)²=0丨x-1丨=0x=1（xy-2)²=0xy=2y=2

7或者-8再问：求过程^_^再答：两个等式两边相加

如果x/y=3/4,求（3x的平方-5xy+2y的平方）/（2x;的平方+3xy-5y的平方因为x/y=3/4,设y=4t（t≠0）,则x=3t所以（3x²-5xy+2y²）/（2

xy/(x+y)=2,1/x+1/y=2(3x-5xy+3y)/(-x+3xy-y)=(3/y+3/x-5)/(-1/y-1/x+3)=(6-5)/(-2+3)=1题目错了,应该是-5xy,不是-5y

x：y=2:3设x=2k,则y=3k∴x²+xy+2y²/x²-xy+y²=(4k²+6k²+9k²)/(4k²-6k&

已知xy+x+y=6则xy+x+y+1=（x+1）（y+1）=7=7×1因为x,y都是自然数（非负整数）所以x+1,y+1≥1所以只能x+1=7y+1=1即x=6,y=0或x+1=1,y+1=7即x=

x^2-xy=14,(1)xy-y^2=-11,(2)(1)-(2)得：x^2-2xy+y^2=14-(-11)=25

（2x+3y-2xy)-(x+4y+xy)-(3xy+2y-2x)=2x+3y-2xy-x-4y-xy-3xy-2y+2x=(2x-x+2x)+(3y-4y-2y)-(2xy+xy+3xy)=3x-3

XY/X+Y=2∴xy=2(x+y)3X-5xY+3Y/-X+3XY-Y=[3(x+y)-5xy]/[-(x+y)+3xy]=[3(x+y)-10(x+y)]/[-(x+y)+6(x+y)]=-7/5

由xy=x+y得y=x/(x-1)>2解不等式,若x>1,则x>2x-2x＜2；若x＜1,则x＜2x-2,x>2,无解.所以x的范围为1＜x＜2.

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