a) Cho biểu thức $\frac{xP}{x+P}-\frac{yP}{y-P}.$ Thay $P=\frac{xy}{x-y}$ vào biểu thức đã cho rồi rút gọn biểu thức.

b) Cho biểu thức $\frac{P^2{Q}^2}{P^2-Q^2}$. Thay $P=\frac{2xy}{x^2-y^2}$ và $Q=\frac{2xy}{x^2+y^2}$ vào biểu thức đã cho rồi rút gọn biểu thức.

Giải

a) Với $P=\frac{xy}{x-y}$

ta có: $\frac{xP}{x+P}-\frac{yP}{y-P}=\frac{{\displaystyle \frac{x^{2}y}{x-y}}}{x+{\displaystyle \frac{xy}{x-y}}}-\frac{{\displaystyle \frac{x{y}^2}{x-y}}}{y-{\displaystyle \frac{xy}{x-y}}}=\frac{x^{2}y}{x^2}-\frac{x{y}^2}{-y^2}=y+x=x+y$

b) Với $P=\frac{2xy}{x^2-y^2}$, $Q=\frac{2xy}{x^2+y^2}$

ta có: $\frac{P^2{Q}^2}{P^2-Q^2}=\frac{{\left({\displaystyle \frac{2xy}{x^2-y^2}}\right)}^2.\left({\displaystyle \frac{2xy}{x^2+y^2}}\right)}{{\left({\displaystyle \frac{2xy}{x^2-y^2}}\right)}^2-{\left({\displaystyle \frac{2xy}{x^2+y^2}}\right)}^2}=\frac{\left[{\displaystyle \frac{2xy.2xy}{\left(x^2-y^2\right)\left(x^2+y^2\right)}}\right]}{{\displaystyle \frac{4x^2{y}^2}{{\left(x^2-y^2\right)}^2}}-{\displaystyle \frac{4x^2{y}^2}{{\left(x^2+y^2\right)}^2}}}$

$$\begin{gathered} =\frac{{\displaystyle \frac{{\left(4x^2{y}^2\right)}^2}{{\left(x^4-y^4\right)}^2}}}{{\displaystyle \frac{4x^2{y}^2\left[{\left(x^2+y^2\right)}^2-{\left(x^2-y^2\right)}^2\right]}{{\left[\left(x^2-y^2\right)\left(x^2+y^2\right)\right]}^2}}}=\frac{{\displaystyle \frac{{\left(4x^2{y}^2\right)}^2}{{\left(x^4-y^4\right)}^2}}}{{\displaystyle \frac{4x^2{y}^2\left(x^4+2x^2{y}^2+y^4-x^4+2x^2{y}^2-y^4\right)}{{\left(x^4-y^4\right)}^2}}} \\ =\frac{{\displaystyle \frac{{\left(4x^2{y}^2\right)}^2}{{\left(x^4-y^4\right)}^2}}}{{\displaystyle \frac{4x^2{y}^2.4x^2{y}^2}{{\left(x^4-y^4\right)}^2}}}=\frac{{\displaystyle \frac{{\left(4x^2{y}^2\right)}^2}{{\left(x^4-y^4\right)}^2}}}{{\displaystyle \frac{{\left(4x^2{y}^2\right)}^2}{{\left(x^4-y^4\right)}^2}}}=1. \end{gathered}$$