Question

Tìm điều kiện của x và y để biểu thức A lớn hơn 1 :

A=\(\left(\dfrac{x}{y^2+xy}-\dfrac{x-y}{x^2+xy}\right)\) : \(\left(\dfrac{y^2}{x^3-xy^2}+\dfrac{1}{x+y}\right):\dfrac{x}{y}\)

Answer 1

A=$\left(\frac{x}{y^2+xy}-\frac{x-y}{x^2+xy}\right)$ : $\left(\frac{y^2}{x^3-x{y}^2}+\frac{1}{x+y}\right):\frac{x}{y}$

A=( \(\dfrac{x}{y\left(x+y\right)}\) - \(\dfrac{x-y}{x\left(x+y\right)}\)) : (\(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}\)+\(\dfrac{1}{x+y}\)) : \(\dfrac{x}{y}\)

A=\(\dfrac{x^2-y\left(x-y\right)}{xy\left(x+y\right)}\) : \(\dfrac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\) : \(\dfrac{x}{y}\)

A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\) : \(\dfrac{y^2-xy+x^2}{x\left(x-y\right)\left(x+y\right)}\):\(\dfrac{x}{y}\)

A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\). \(\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2-xy+y^2}\):\(\dfrac{x}{y}\)

A = \(\dfrac{x-y}{y}\) : \(\dfrac{x}{y}\)

A = \(\dfrac{x-y}{x}\)

A= 1 - \(\dfrac{y}{x}\)>1

=> y/x <0

=> xy<0 , x+y khác 0