x y=12_xy=9则x-y等于-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

是不是求(x2＋3xy＋2y2)／(x2y+2xy2）,如果是,则[（x+y)2+y(x+y)]／[xy(x+2y)]再化为（x+y)／xy=4／3

1、平方差原式=[(2x+3y)+(x+2y)][(2x+3y)-(x+2y)]=(2x+3y+x+2y)(2x+3y-x-2y)=(3x+5y)(x+y)2、3|x-3|+(x-3y)²=

x的平方+3xy+2y的平方除以x的平方y+2xy的平方=(x+2y)(x+y)/xy(x+2y)=12/9=4/3再问：那个再问一下不改变分式的值使分式的分子分母首项系数都是正数-(负7xy/负8z

你确定没写错题目?后面的式子一约就变成前面的式子了,答案也是8/xy再问：怎么个约法?再答：y/xyyy约掉剩下1/x

必要性：若u=fg则u'x=f'gu'y=fg'u"xy=f'g'所以uu"xy=fg*f'g'=fg'*f'g=u'x*u'y必要性成立充分性：若uu"xy=u'x*u'yuu"xy-u'x*u'y

（4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22

画出限定区域z=2x-yy=2x-z当y=2x-z经过点C时,z有最大值z最大值=2*2-（-1）=5如果您认可我的回答,请点击“采纳为满意答案”,祝学习进步!再问：�Ͽ��Ͽɣ��Щ֪ʶ�Ǹ��

x平方+3xy+y平方=x²-2xy+y²+5xy=(x-y)²+5xy=(-3)²+5*9=9+45=54

将xy=2x-2y带入3x-4xy-3y即可得出-2/5

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

因为xy/（x+y）=1/2所以x+y=2xy原式=3（x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

-2x+6xy-2y/(7x+5y)-9xy-(3x+y)=[3(x+y)-x-y]/[7x+5y-9/2(x+y)-3x-y]=2(x+y)/[-1/2(x+y)]=-4

xy/(x+y)=2(x+y)/xy=1/21/x+1/y=1/2(3x-xy+3y)/(-x+3xy-y)=(3/y-1+3/x)/(-1/y+3-1/x)=[3(1/x+1/y)-1]/[(3-(

lg(xy)=0xy=1则:|x|＝1而Y=X,|x|不等于Y所以x=-1,y=-1原式=log8（2）=1/3

x²+xy=12 xy+y²=13

两式相加得到x+y=5,相减得y-x=1/5,故x=12/5,y=13/5xy=156/25,因为要求的都是正数,而且xy同正负,所以只考虑x,y正数即可故x²+y²=（x+y）^

16524

XY/（X+Y）=3,XY=3（X+Y）原式=［6（X+Y）+4XY］/［9（X+Y）+4XY］=［6（X+Y）+12（X+Y）］/［9（X+Y）+12（X+Y）］=［18（X+Y）］/［21（X+Y

(x^2+3xy+2y^2)÷(x^2y+2xy^2)=(x+2y)(x+y)/xy(x+2y)=(x+y)/xy=12/9=4/3

因为,x-y=3xy所以：-3x+6xy+3y/[(7x-5y)-9xy-(3x-y)]=[-3(x-y)+6xy]/[(7x-5y)-9xy-(3x-y)]=[-3*3xy+6xy]/[7x-5y-