Luyện tập 5 trang 42 SGK Toán 8 Tập 1 Cánh diều -CD

Ta có: \(\frac{x-5y}{2x-3y}-\frac{24xy}{4{{x}^{2}}-9{{y}^{2}}}-\frac{x+8y}{3y-2x}\)

\(\begin{align} & =\frac{x-5y}{2x-3y}-\frac{24xy}{(2x+3y)(2x-3y)}+\frac{x+8y}{2x-3y} \\ & =\frac{x-5y}{2x-3y}+\frac{x+8y}{2x-3y}-\frac{24xy}{(2x+3y)(2x-3y)} \\ & =\frac{x-5y+x+8y}{2x-3y}-\frac{24xy}{(2x+3y)(2x-3y)} \\ & =\frac{2x+3y}{2x-3y}-\frac{24xy}{(2x+3y)(2x-3y)} \\ & =\frac{{{(2x+3y)}^{2}}}{(2x+3y)(2x-3y)}-\frac{24xy}{(2x+3y)(2x-3y)} \\ & =\frac{{{(2x+3y)}^{2}}-24xy}{(2x+3y)(2x-3y)}=\frac{4{{x}^{2}}+12xy+9{{y}^{2}}-24xy}{(2x+3y)(2x-3y)} \\ & =\frac{4{{x}^{2}}-12xy+9{{y}^{2}}}{(2x+3y)(2x-3y)}=\frac{{{(2x-3y)}^{2}}}{(2x+3y)(2x-3y)}=\frac{2x-3y}{2x+3y} \\ \end{align}\)

-- Mod Toán 8 HỌC247