已知x y=30,x-y=20,求1/5(x-3y)的平方-12
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由X+Y=3XY可得X+Y-XY=2XY,用X+Y-XY替换分母中的2XY,所以(2X+2Y-XY)/(X+Y+2XY)=(2X+2Y-XY)/(2X+2Y-XY)=1
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)-(xy-y)=40-(-20)x-xy-xy+y=40+20x+y-2xy=60(x-xy)+(xy-y)=40+(-20)x-xy+xy-y=40-20x-y=20
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+y=3xy=3(x+y)3x-5xy+3y/-x+3xy-y=(3x+3y)-5*3(x+y)/[-x-y+3*3(x+y)]=-12(x+y)/8(x+y)=-3/2望采纳,谢谢!
y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5
-xy(x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10
xy/(x+y)=2∴xy=2(x+y)(3xy-x-y)/(3x-5xy+3y)=[6(x+y)-x-y]/[3x+3y-10(x+y)]=5(x+y)/[-7(x+y)]=-5/7
x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60希望满意!
(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8
2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3(x+y)+2xy=3*(-2)+2*3=0
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
xy+1/xy>=2√(xy*1/xy)=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√(x/y*y/x)=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求
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
3(x^2-xy)-xy+y^2=3(x^2-xy)-(xy-y^2)=3*5-(-3)=15+3=18
原式=-xy²(x²y^4-xy²-1)∵xy²=-2原式=2((-2)²-(-2)-1)=10
x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60