Cho biểu thức (xP)/(x+P) − (yP)/(y−P)

Giải Toán 8 Bài Ôn tập chương 2

Video Giải Bài 59 trang 62 Toán 8 Tập 1

Bài 59 trang 62 Toán 8 Tập 1:

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

Lời giải

a) Thay $P=\frac{xy}{x-y}$ vào biểu thức ta được: $\frac{x.\frac{xy}{x-y}}{x+\frac{xy}{x-y}}-\frac{y.\frac{xy}{x-y}}{y-\frac{xy}{x-y}}$

Ta có: 

$\begin{array}{l}\frac{x.\frac{xy}{x-y}}{x+\frac{xy}{x-y}}-\frac{y.\frac{xy}{x-y}}{y-\frac{xy}{x-y}}\\ =\frac{\frac{x^{2}y}{x-y}}{\frac{x\left(x-y\right)}{x-y}+\frac{xy}{x-y}}-\frac{\frac{x{y}^2}{x-y}}{\frac{y\left(x-y\right)}{x-y}-\frac{xy}{x-y}}\\ =\frac{\frac{x^{2}y}{x-y}}{\frac{x^2-xy+xy}{x-y}}-\frac{\frac{x{y}^2}{x-y}}{\frac{xy-y^2-xy}{x-y}}\\ =\frac{\frac{x^{2}y}{x-y}}{\frac{x^2}{x-y}}-\frac{\frac{x{y}^2}{x-y}}{\frac{-y^2}{x-y}}\\ =\left(\frac{x^{2}y}{x-y}:\frac{x^2}{x-y}\right)-\left(\frac{x{y}^2}{x-y}:\frac{-y^2}{x-y}\right)\\ =\left(\frac{x^{2}y}{x-y}.\frac{x-y}{x^2}\right)-\left(\frac{x{y}^2}{x-y}.\frac{x-y}{-y^2}\right)\\ =y-\left(-x\right)=y+x\end{array}$

b) Thay $P=\frac{2xy}{x^2-y^2}$ và $Q=\frac{2xy}{x^2+y^2}$ vào biểu thức trên ta được:

$\begin{array}{l}\frac{{\left(\frac{2xy}{x^2-y^2}\right)}^2.{\left(\frac{2xy}{x^2+y^2}\right)}^2}{{\left(\frac{2xy}{x^2-y^2}\right)}^2-{\left(\frac{2xy}{x^2+y^2}\right)}^2}\\ =\frac{{\left(\frac{2xy}{x^2-y^2}.\frac{2xy}{x^2+y^2}\right)}^2}{\left(\frac{2xy}{x^2-y^2}-\frac{2xy}{x^2+y^2}\right)\left(\frac{2xy}{x^2-y^2}+\frac{2xy}{x^2+y^2}\right)}\\ =\frac{{\left.\frac{2xy.2xy}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right]}^2}{\left(\frac{2xy}{x^2-y^2}-\frac{2xy}{x^2+y^2}\right)\left(\frac{2xy}{x^2-y^2}+\frac{2xy}{x^2+y^2}\right)}\\ =\frac{{\left.\frac{4x^2{y}^2}{x^4-y^4}\right]}^2}{\left.\frac{2xy\left(x^2+y^2\right)-2xy\left(x^2-y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right]\left.\frac{2xy\left(x^2+y^2\right)+2xy\left(x^2-y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right]}\\ =\frac{\frac{14x^4{y}^4}{\left(x^4-y^4\right)}}{\left(\frac{2x^{3}y+2x{y}^3-2x^{3}y+2x{y}^3}{x^4-y^4}\right)\left(\frac{2x^{3}y+2x{y}^3+2x^{3}y-2x{y}^3}{x^4-y^4}\right)}\\ =\frac{\frac{16x^4{y}^4}{\left(x^4-y^4\right)}}{\left(\frac{4x{y}^3}{x^4-y^4}\right)\left(\frac{4x^{3}y}{x^4-y^4}\right)}\\ =\frac{\frac{16x^4{y}^4}{\left(x^4-y^4\right)}}{\frac{16x^4{y}^4}{{\left(x^4-y^4\right)}^2}}=1\end{array}$