a, \(\frac{x+4}{x^2-1}-\frac{x-5}{x^2-x}\)
\(=\frac{x+4}{\left(x-1\right)\left(x+1\right)}-\frac{x-5}{x\left(x-1\right)}\)
\(=\frac{\left(x+4\right)\left(x^2-x\right)-\left(x-5\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1-x\right)}\)
\(=\frac{\left(x^3+4x^2-x^2-4x\right)-\left(x^3-5x^2-x+5\right)}{x^2-x+x-1-x^2+x}\)
\(=\frac{x^3+4x^2-x^2-4x-x^3-5x^2-x+5}{x-1}\)
\(=\frac{-2x^2-5x+5}{x-1}=\frac{-2x^2}{x-1}-\frac{5\left(x-1\right)}{\left(x-1\right)}=\frac{-2x^2}{x-1}-5\)
b) \(\frac{6x+10}{y^2-25}:\frac{3x+5}{5y-25}\)
\(=\frac{6x+10}{\left(y-5\right).\left(y+5\right)}.\frac{3x+5}{5.\left(y-5\right)}\)
\(=\frac{\left(6x+10\right).5.\left(y-5\right)}{\left(y-5\right).\left(y+5\right).\left(3x+5\right)}\)
\(=\frac{2.\left(3x+5\right).5.\left(y-5\right)}{\left(y-5\right).\left(y+5\right).\left(3x+5\right)}\)
\(=\frac{2.5}{y+5}=\frac{10}{y+5}\)
Vậy : \(\frac{6x+10}{y^2-25}:\frac{3x+5}{5y-25}=\frac{10}{y+5}\)
