已知x² 5y²-4xy² 2y 1等于0

设,y=k/x+a/x,然后分别把两个值带入,解方程组k=xy=-2,和交点带入第二个式子求m,再联立解方程组.

,y1与x-1成正比例,y2与x正比例可设y1=k(x-1),y2=tx则y=y1-y2=k(x-1)-tx当X=2时,y=4;当x=-1时,y=-5则4=k-2t-5=-2k+t解得k=2,t=-1

y1=k1x,y2=k2/x^2y=k1x+k2/x^2代入数值-4=k1+k25=2k1+k2/4k1=24/7,k2=-52/7y=24/7x-52/7x^2

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

直接降维呗y2=y1*u=xcosxuy'=(cosx-xsinx)u+xcosxu'y''=(-sinx-sinx-xcosx)u+(cosx-xsinx)u'+(cosx-xsinx)u'+xco

5x+4y-3x-2y+xy=2x+2y+xy=2(x+y)+xy因：x+y=5,xy=2所以原式=2X5+2=12

x²+y²-2x+4y+5=0配方：(x-1)²+(y+2)²=0所以只有x-1=0且y+2=0∴x=1,y=-2∴(X^4-y^4)/(2x^2+xy-y&#

根据已知得到x=-2y=-1原式=5xy^2-{2x^2y-[3xy^2-(4xy^2-2x^2y)]}=5xy^2-{2x^2y-[3xy^2-4xy^2+2x^2y]}=5xy^2-{2x^2y-

原式=[4(x+y)-2xy]分之[(x+y)+xy]=[4(3xy)-2xy]分之[(3xy)+xy]=10xy分之2xy=5分之1

解题思路：根据题意，由完全平方公式的知识可求解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/inc

x+根号xy-6y=0(根号x-2根号y)(根号x+3根号y)=0解得根号x=2根号y即x=4y或根号y+3根号y=0不成立所以4x+5根号xy-6y分之x+2根号xy=[4y+2根号(4y*y)]/

（x+2）^2+|x+y+5|=0,x+2=0,x+y+5=0x=-2,y=-33(x^2)y-2(x^2)y+2xy-(x^2)y+4x^2-xy=4x^2+xy=4*(-2)^2+(-2)*(-3

y=ax^2+b/(x^1/2)5=a+b18=16a+b/2解得a=1,b=4.y=x^2+4/(x^1/2)

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

已知A=5x^2y-3xy^2+4xy,B=7xy^2-2xy+x^2y.（1）求A-2B的值A-2B=5x^2y-3xy^2+4xy-2*(7xy^2-2xy+x^2y)=5x^2y-3xy^2+4

5x+4y-3x-2y+xy=2x+2y+xy=2(x+y)+xy因：x+y=5,xy=2所以原式=2X5+2=12你抄错了应该是5x+4y-3x-2y+xy

2x＋y＋3=xy,①若x=1,则y不存在,此时5x＋4y无最值；②若x≠1,则y=(2x＋3)/(x－1),M=5x＋4y=5x＋(8x＋12)/(x－1)=(5x²＋3x＋12)/(x－

x+2=0,x+y+5=0x=-2,y=-33x^2y+{-2X^2y-[-2xy+(x^2y-4x^2)]-xy}=3x^2y-2X^2y+2xy-x^2y+4x^2-xy=4x^2+xy=4*4+

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

x^2+y^2-2x+4y+5=0(x-1)^2+(y+2)^2=0x=1,y=-2化解式子,再代入