从xy=1能否得到x=y分之一为什么-搜答案-搜狗做题网

(x^2)/2+xy+(y^2)/2=1/2*(x^2+2xy+y^2)=(x+y)^2/2=1/2

x-3xy-Y分之3x+5xy-3y==(3x+5xy-3y)/(x-3xy-Y)=(5+3*(1/y-1/x))/(-3+(1/y-1/x)).分子分母同时除以xy=(5-21)/(-3-7).将x

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-2,前提是xy都不为零再问：能说一下解题步骤吗？

xy=x-y两边同时除以xy1=1/y-1/xy分之一减x分之一等于1

1/2x²+xy+1/2y²=1/2(x²+2xy+y²)=1/2(x+y)²=1/2

|x-1|+|y+3|=0|x-1|≥0|y+3|≥0所以x-1=0,y+3=0x=1,y=-3原式=1/1*-3+1/3*-1+1/1*5+1/3*7+1/5*9+.+1/(2n-1)(2n+3)=

由已知得：xy=-1,1/x-1/y=(y-x)/(xy)=5那么：y-x=5xy=-5所以：(y-x)²=25即：y²-2xy+x²=25得：y²+x

1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+…+1/(x+2010)(y+2010)=1/（1*2）+1/（2*3）+1/（3*4）+.+1/（2011*2012)=1/1-1/2+

xy=x-y,因为求1/x-1/y,则x,y不会是0,所以等式两边同时处以xy,xy/xy=(x-y)/xy,1=1/y-1/x,则1/x-1/y=-1

x^(y-1)=yx=y^[1/(y-1)]底和幂无法合并x^y=xy不能给出表达式f(x)

∵|X-1|+(XY-2)的2次方=0∴x-1=0xy-2=0x=1y=2∴xy分之一+（x+1）（y+1）分之一+（x+2）（y+2）分之一+…+（x+2011）（y+2011）分之一=(1*2)分

因为(x+1/y)(y+1/x)=xy+1+1+1/xy=18+2=20所以y+1/x=20/5=4再问：哦再问：我没想到，，，，

1/x+1/y=（x+y)/xy=5x+y=5xy所以原式=[2(x+y)-3xy]÷[(x+y)+2xy]=[2(5xy)-3xy]÷[(5xy)+2xy]=7xy÷7xy=1再问：1/x+1/y=

1/x+1/y=(x+y)/xy=5==>x+y=5xy所以2x-3xy+2y=2(x+y)-3xy=2*5xy-3xy=7xyx+2xy+y=5xy+2xy=7xy所以2x-3xy+2y除以x+2x

3x-3y=xy3/x-3/y=1(两边同除xy)1/x-1/y=1/3

x^2y+xy^2=xy(x+y)=1/5(x^2+1)(y^2+1)=(xy)^2+x^2+y^2+1=(xy)^2+(x+y)^2-2xy+1=1/25+1-2/5+1=41/25再问：（X的平方

1/x-1/y=2(y-x)/xy=3y-x=3xyx-y=-3xy原式=[2(x-y)-14xy]/[(x-y)-2xy]=[2(-3xy)-14xy]/[(-3xy)-2xy]=-20xy/(-5

1/x-1/y=(y-x)/xy=-(x-y)/(x-y)=-1

x:y=8:9设x=8ky=9k1/x+1/y=1/8k+1/9k=9/72k+8/72k=17/72k=1/72k/17=1/21672k/17=216k=51x+y=17k=51*17=867