Luyện tập 4 trang 41 SGK Toán 8 Tập 1 Cánh diều - CD

a) Ta có: \(\frac{4x+3y}{{{x}^{2}}-{{y}^{2}}}-\frac{3x+4y}{{{x}^{2}}-{{y}^{2}}}\)

\(\begin{align} & =\frac{4x+3y-3x-4y}{(x+y)(x-y)}=\frac{(4x-3y)+(3y-4y)}{(x+y)(x-y)} \\ & =\frac{x-y}{(x+y)(x-y)}=\frac{1}{x+y}=\frac{x-y+2y}{x+y} \\ & =\frac{x+y}{x+y}=1 \\ \end{align}\)

b) Ta có: \(\frac{2xy+3{{y}^{2}}}{{{x}^{2}}-3xy}-\frac{x}{3x-9y}\)

\(\begin{align} & =\frac{y(2x-3y)}{x(x-3y)}-\frac{x}{3(x-3y)} \\ & =\frac{3y(2x-3y)}{3x(x-3y)}-\frac{{{x}^{2}}}{3x(x-3y)} \\ & =\frac{6xy-9{{y}^{2}}-{{x}^{2}}}{3x(x-3y)}=\frac{{{x}^{2}}-6xy+{{(3y)}^{2}}}{3x(x-3y)} \\ & =-\frac{{{(x-3y)}^{2}}}{3x(x-3y)}=-\frac{x-3y}{3x} \\ & =\frac{3y-x}{3x} \\ \end{align}\)

-- Mod Toán 8 HỌC247