已知x-y=5,xy=2,求2x²+2y²-1的值-搜答案-搜搜考试网

xy/x+y=3则xy=3(x+y)2x-5xy+2y/x-3xy+y=2(x+y)-5xy/(x+y)-3xy=2(x+y)-15(x+y)/(x+y)-9(x+y)=-13(x+y)/-8(x+y

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

x-y=2xy3x-5xy-3y/x+xy-y=[3(x-y)-5xy]/[(x-y)+xy]=(6xy-5xy)/(2xy+xy)=1/31/x+1/y=4y+x=4xyx-5xy+y/2x+xy+

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

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

答：x+y=-1,xy=-2-5（x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)

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

因为X+Y分之XY=2,所以XY=2(X+Y)代入后面的分子式得：13（X+Y)/5(X+Y)=13/5

化简已知条件得：xy=2(x+y)代入算式,得：（3(x+y)-10(x+Y))/(6(x+Y)-(x+y))＝-7/5.注：算式中的5Y应为5XY吧

-5（x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4（x+y）+2xy=-4×（-1）+2×（-2）=4+（-4）=0你有问题也可以在这里向我提问

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.

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

x^3y-2x^y^2+xy^3=xy(x^2+2xy+y^2-4xy)=xy[(x+y)^2-4xy]=3*(5^2-4*3)=3*13=39

这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-

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

xy/(x+y)=2xy=2(x+y)(3x-5xy+3y)/(-x+2xy-y)=[3(x+y)-5xy]/[2xy-(x+y)]=[3(x+y)-10(x+y)]/[4(x+y)-(x+y)]=-

xy=2(x+y)x+y=xy/2原式=[3(x+y)-5xy]/[-(x+y)+3xy]=(3xy/2-5xy)/(-xy/2+3xy)=(-7xy/2)/(5xy/2)=-7/5

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－

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