,A为y＝-x分之1 x分之2,三角形AOB面积怎么变化-搜答案-神马作文网

原式=[(x+2y)/(x+y)][xy/(x+2y)]÷[(x+y)/xy]=[xy/(x+y)]×xy/(x+y)=x²y²/(x+y)²

由1/x-1/y=3得(y-x)/xy=3题中的式子分子分母同时除以xy得(2/y+3-2/x)/(1/y-2-1/x),通分得【2(x-y)/xy+3】/【(x-y)/xy-2】代入得3/5

x/a+y/b=1x=1,y=4:====>1/a+4/b=1.(1)x=2,y=10:===>2/a+10/b=1.(2)(1)*2-(2):-2/b=1===>1/b=-1/21/a=1-4/b=

1、已知a+b分之1=a分之1+b分之1,通分整理得（a+b)^2=ab,即a^2+b^2=ab所以所求的通分后为1答案为:1

1 x^2y分之x^4

1x^2y分之x^4=x²/y2ab分之ab-b^2(b不等于0）=1-b/a36abc^2分之36ab^3c=6b²/c4a^2-b^2分之（a+b)^3=（a+b）²

由定义域确定x,y1-4x>=0,4x-1>=0==>x=1/4==>y=1/2.略

1/x-1/y=(y-x)/xy=3y-x=3xy所以x-y=-3xy所以原式=[(x-y)-2xy]分之[2(x-y)-5xy]=[(-3xy)-2xy]分之[2(-3xy)-5xy]=(-5xy)

(2x-3)/(x^2-x)=(2x-3)/x(x-1)A/(x-1)+B/x=(Ax+Bx-B)/x(x-1)=[(A+B)x-B]/x(x-1)两式对比A+B=2B=3A=2-B=-1所以A+B=

原式=1/2x-1/(x+y)[(x+y)/2x-(x+y)]=1/2x-(1/2x-1)=1/2x-1/2x+1=1

原式=[(x+2y)/(x+y)]×[xy/(x+2y)]÷[(x+y)/xy]=[(x+2y)/(x+y)]×[xy/(x+2y)]×[xy/(x+y)]=x²y²(x+2y)/

（1）=x平方-y平房分之x(x+y)-y(x-y)+2xy=（x+y）*（x-y）分之（x+y）的平方=x-y分之x+y（2）=a-1分之2-a-1分之a+1=-1

右边=[a(x-1)-b(3x+2)]/(x-1)(3x+2)=[(a-3b)x+(-a-2b)]/(3x²-x-2)=(4x-9)/(3x²-x-2)所以(a-3b)x+(-a-

Y=√(1-4X)+√(4X-1)+1/2,即1-4x≥04x-1≥0,即1-4x=0x=1/4,y=1/2√(x/y+2+y/x)-√(x/y-2+y/x)=√(1/2+2+2)-√(1/2-2+2

(1)=x/y(2)=x/4

（1）2x平方y分之1+3x平方分之2-4xy平方分之3=6y/(12x^2y^2)+8y^2/(12x^2y^2)-9x/(12x^2y^2)=(6y+8y^2-9x)/(12x^2y^2)（2）x

是这样的形式吗?[x/（x+y）+2y/（x+y）]×[xy/（x+y）]÷（1/x+1/y）一:原式=（x+2y）/（x+y）×[xy/（x+2y）]÷[（x+y）/xy]=x²y&sup

分式有：a分之1、6+x分之5、7分之x+8分之y、9x+y分之10共4个