已知x y=30,x-y=20,求1/5(x-3y)的平方-12-搜答案-搜搜做题作业网

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

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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