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