Bài 1:
a) Ta có: \(\left(12x^3-28x^2+21x-5\right):\left(6x-5\right)-\left(2x^2-4x\right)\)
\(=\left(12x^3-10x^2-18x^2+15x+6x-5\right):\left(6x-5\right)-\left(2x^2-4x\right)\)
\(=\frac{2x^2\left(6x-5\right)-3x\left(6x-5\right)+\left(6x-5\right)}{6x-5}-2x^2+4x\)
\(=\frac{\left(6x-5\right)\left(2x^2-3x+1\right)}{6x-5}-2x^2+4x\)
\(=2x^2-3x+1-2x^2+4x\)
\(=x+1\)
b) Ta có: \(\left(\frac{x+1}{x-3}+\frac{5x-39}{x^2-9}-\frac{11}{x+3}\right):\frac{x^2+2x+1}{2x+6}\)
\(=\left(\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{5x-39}{\left(x-3\right)\left(x+3\right)}-\frac{11\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{2\left(x+3\right)}{\left(x+1\right)^2}\)
\(=\frac{x^2+4x+3+5x-39-11x+33}{\left(x+3\right)\left(x-3\right)}\cdot\frac{2\left(x+3\right)}{\left(x+1\right)^2}\)
\(=\frac{x^2-2x-3}{x-3}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{x^2-3x+x-3}{x-3}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{x\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}\cdot\frac{2}{\left(x+1\right)^2}\)
\(=\frac{\left(x-3\right)\left(x+1\right)\cdot2}{\left(x-3\right)\left(x+1\right)^2}\)
\(=\frac{2}{x+1}\)
