9.先阅读下列解法,再解答后面的问题.
已知$\frac{3x-4}{{x}^{2}-3x+2}$=$\frac{A}{x-1}$+$\frac{B}{x-2}$,求A、B的值.
解法一:将等号右边通分,再去分母,得:3x-4=A(x-2)+B(x-1),
即:3x-4=(A+B)x-(2A+B),
∴$\left\{\begin{array}{l}A+B=3\\-(2A+B)=-4\end{array}\right.$.
解得 $\left\{\begin{array}{l}A=1\\ B=2\end{array}\right.$.
解法二:在已知等式中取x=0,有-A+$\frac{B}{-2}$=-2,整理得
2A+B=4;
取x=3,有$\frac{A}{2}$+B=$\frac{5}{2}$,整理得
A+2B=5.
解 $\left\{\begin{array}{l}2A+B=4\\ A+2B=5\end{array}\right.$,
得:$\left\{\begin{array}{l}A=1\\ B=2\end{array}\right.$.
(1)已知$\frac{11x}{{-3{x^2}-14x+24}}=\frac{A}{x+6}+\frac{B}{4-3x}$,用上面的解法一或解法二求A、B的值.
(2)计算:
[$\frac{1}{{(x-1)({x+1})}}+\frac{1}{(x+1)(x+3)}+\frac{1}{(x+3)(x+5)}+…+\frac{1}{(x+9)(x+11)}$](x+11),并求x取何整数时,这个式子的值为正整数.