计算:(x^-1+y^-1)^2*(1/x^2+2xy+y^2)^-1
1个回答
(x^-1+y^-1)^2*(1/x^2+2xy+y^2)^-1
=[(x+y)/(xy)]²*[1/(x+y)²]^(-1)
=(x+y)²/(xy)² *(x+y)²
=(x+y)^4/(xy)²
相关问题
计算(1/2xy+y²+1)+(x²-1/2xy-2y²-1)
计算:(1/x-y-1/x+y)/2y/x^2-2xy+y^2,x=1+根号2,y=1-根号2
计算(3xy-2x²+1/2y²)(3xy+2x²-1/2y²)
计算(3x^3y-x^2y+1/2xy)÷(-1/2xy)
计算(1)(−2xy2)2÷(−14x2y2)•4xy(2)(3x-2y+1)(3x+2y-1)
计算:(1)(√xy+2√y/x+√x/y+√1/xy)√xy (2)(√5-√3)^2-(√5+√3)^2
计算:1.(x-1)^2+2(1-x) 2.(x-y)/(x+3y)÷(x^2-y^2)/(x^2+6xy+9y^2)-
计算(2x-1)(x²-xy+2y²)
计算(2/5x²y)³*(1/2x²y)²÷(-1/5xy)³
[2/xy/(1/x+1/y)2+(x2+y2)/(x2+2xy+y2)]*2x/(x-y)