已知绝对值XY-2与绝对值Y-1互为相反数试求XY分之1-搜答案-大师作文网

1:绝对值永远大于等于0|xy-2|+|y-1|=0|xy-2|=0,|y-1|=0xy=2y-1=0y=1,x=22:1/xy+1/(x+1)(y+1)+...+1/(x+2009)(y+2009)

已知x-1的绝对值+y-2的绝对值=0x-1=0、y-2=0x=1、y=22x^2-[x^2-(2xy-y^2)+2xy]=2-[1-(4-4)+4]=-3

|x-3|+（Y+2）^2=0|x-3|=0（Y+2）^2=0x-3=0x=3Y+2=0y=-2y^2-xy=(-2)^2-3*(-2)=4+6=10

根据绝对值的性质得x-1=0,y-2=0x=1,y=2xy=2

当x=2y=3时,x-y+xy=5当x=2y=-3时,x-y+xy=-1当x=-2y=3时,x-y+xy=-11当x=-2y=-3时,x-y+xy=7

x+3y+6=02y-y+5=0x=9y=-5(x+y)-xy=4-(9*-5)=49希望能够帮上你!

X-3绝对值+Y+2绝对值=0X-3绝对值=0X-3=0X=3Y+2绝对值=0Y+2=0Y=-2x+y/xy=1/-6

3x+2y+1绝对值+(3x-4y+7)²=0所以3x+2y+1=03x-4y+7=0相减6y-6=0所以y=1x=(4y-7)/3=-1所以xy=-1

x=1y=1代入就好了

依题意有：√(x+3y-5)+|2x-y-3|=0而,√(x+3y-5)≥0,|2x-y-3|≥0所以,只有两者同时为零时它们的和才为零即,x+3y-5=0,2x-y-3=0联立解得：x=2,y=1所

x-1的绝对值+y-3的绝对值=0,绝对值都大于等于0相加为0则各项均为0所以x-1=0y-3=0所以x=1,y=3所以xy/x+y=3/4

由题可知x=-3,y=-5,x+y=-8,

xy>0则x和y同号所以x=2,y=3或x=-2,y=-3所以x+y=5或-5|x+y|=5x-y=-1或1|x-y|=1所以原式=5-1=4再问：你确定再答：嗯

∵│xy-2│+│y-1│=0.∴y-1=0,即y=1;xy-2=0,x*1-2=0,x=2.则2/xy+2/(x+1)(y+1)+2/(x+2)(y+2)+…+2/(x+2012)(y+2012)=

∵|x-y+2|与|x+y-1|为相反数∴x-y+2=x+y-1=0∴x=-1/2y=3/2∴xy=-3/4∴xy负倒数为4/3

|x+1|>=0,|y-2|>=0,又它们互为相反数,即x+1=y-2=0,所以|x|+|y|=3

|x-y|=4,xy=-3(x+y)^2=(x-y)^2+4xy=14-12=2x+y=根2,-根2xy^2+x^2y=xy(x+y)=-3根2,3根2再问：(x+y)^2=(x-y)^2+4xy=1

即|x-1|+|y-2|=0所以x-1=0,y-2=0x=1,y=2所以|x|+|y|=1+2=3再问：你确定么再答：确定

正负12..正负1正负10负2013再问：我想追问再答：恩客气了