已知x-y=5,xy=2,求2x²+2y²-1的值
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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