Tính giá trị \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\) .
\(-2\).\(2\).\(1\).\(-1\).Hướng dẫn giải:Ta có \(\left(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}\right)=\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}=\frac{1}{x-2}\left(\frac{1}{x-1}+\frac{1}{x-3}\right)=\frac{2x-4}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
nên giới hạn cần tính bằng \(\frac{2}{\left(2-1\right)\left(2-3\right)}=-2.\)