a) \(\frac{2x^3+x^2+x+6}{x^2-x+2}=\frac{\left(2x+3\right)\left(x^2-x+2\right)}{x^2-x+2}=2x+3\)

b) \(\frac{x}{x-3}-\frac{5x^2+27}{x^2-9}+\frac{x-9}{x+3}\)

\(=\frac{x}{x-3}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x-9}{x+3}\)

\(=\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-9\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-12x+27}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+3x-\left(5x^2+27\right)+x^3-12x+27}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-3x^2-9x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-3x}{x-3}\)

Với a\(\in\)Z thì a³-a=(a-1)a(a+1) là tích 3 số tự nhiên liên tiếp nên chia hết cho 2,3

Mà (2,3)=1 => a³-a chia hết cho 6

=> S-P=(a₁³-a₁)+(a₂³-a₂)+....+(a_n³-a_n) chia hết cho 6

Vậy S chia hết cho 6 <=> P chia hết cho 6

\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow\frac{x-2}{x+2}+\frac{3}{x-2}-\frac{x^2-11}{x^2-4}=0\)

<=> \(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}+\frac{3x+6}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{x^2-4x+4+3x+6-x^2+11}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{-x+21}{\left(x-2\right)\left(x+2\right)}=0\)

=> -x+21=0

<=> -x=-21

<=> x=21 (tmđk)

Vậy x=21 là nghiệm của pt

\(\frac{x}{2x-6}-\frac{2}{2x+2}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne-1;x\ne3\right)\)

<=> \(\frac{x}{2x-6}-\frac{2}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

<=> \(\frac{x}{2\left(x-3\right)}-\frac{2}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

<=> \(\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-3\right)}-\frac{2\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{\left(x+1\right)\left(x-3\right)2}=0\)

<=> \(\frac{x^2+2x+1}{2\left(x+1\right)\left(x-3\right)}-\frac{2x-6}{2\left(x+1\right)\left(x-3\right)}-\frac{4x}{2\left(x+1\right)\left(x-3\right)}=0\)

<=> \(\frac{x^2+2x+1-2x-6-4x}{2\left(x+1\right)\left(x-3\right)}=0\)

<=> \(\frac{x^2-4x-5}{2\left(x+1\right)\left(x-3\right)}=0\)

=> x²-4x-5=0

<=> x²-5x+x-5=0

<=> x(x-5)+(x-5)=0

<=> (x-5)(x+1)=0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

Đối chiếu điều kiện => x=5

Vậy x=5 là nghiệm của pt

\(C=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x}{x-1}-\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)\)

\(=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-1\right)\left(x+3\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{3x^2-x+2-x^2-3x-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}=\frac{x-1}{x+3}\)

Vậy C=\(\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)\)