Câu 1:
a: \(=\dfrac{6x^2-2x-9x^2-3x}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{-3x^2-5x}{\left(3x-1\right)\left(3x+1\right)}\)
b: \(=\dfrac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{10x-25}{x^3-25x}\)
c: \(=\dfrac{-32x^2}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2x-1}{x\left(2x+1\right)}+\dfrac{2x+1}{x\left(2x-1\right)}\)
\(=\dfrac{-32x^3-4x^2+4x-1+4x^2+4x+1}{x\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{-32x^3+8x}{x\left(2x+1\right)\left(2x-1\right)}=\dfrac{-8x\left(4x^2-1\right)}{x\left(2x+1\right)\left(2x-1\right)}=-8\)
d: \(=\dfrac{-2x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x^2+x+1}\)
\(=\dfrac{-2x^2+x^2+x+1+x^2-x}{x\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x\left(x-1\right)\left(x^2+x+1\right)}\)
