lim(x,y)→(0,0),(根号(1+xy)-1)/根号(x²+y²)=?
证明lim(x,y)→(0,0),xy/根号(x²+y²)=0
求极限lim x→0 y→0 2xy/根号下1+xy 然后-1 {不在根号里}
lim (x,y)->(0,0) xy/[根号下(xy+1)]-1的值为
lim (x,y)->(0,0) xy/[根号下(xy+1)]-1的值为
已知x平方+y平方-2y+1=0,求(根号x+根号y+3)/((根号xy+y)(3根号x+根号y))
求极限:lim xy分之{[(1+xy)开3次根号]-1},(x,y)→(0,0)
lim(x,y)-(0,0)=根号下(xy+9)-3/xy
已知x、y满足根号下4x-5y+根号下x-y-1=0,则根号下xy-根号下x/y
化简求值:(X×根号X+X×根号y)÷(XY-Y²)-(X+根号XY+Y)÷X×根号X-Y×根号Y,其中X=1
根号2x-y-8+根号x+2y+1=0 求xy.
先化简,在求值:(x+根号xy)/(y+根号xy)+((根号xy)-y)/(x-根号xy),其中x=根号3+1,y=根号
已知y=根号1-x+根号x-1+3,求根号x+根号y分之x+2根号xy+y+根号x-根号y分之一的值