Giá trị của biểu thức \(A=\dfrac{x-6}{x^2+1}.\dfrac{3x^2-3x+3}{x^2-36}+\dfrac{x-6}{x^2+1}.\dfrac{3x}{x^2-36}\) tại \(x=994\) là
\(\dfrac{3}{988}.\)\(\dfrac{3}{1000}.\)\(\dfrac{1}{1000}.\)\(\dfrac{1}{988}.\)Hướng dẫn giải:Ta có \(A=\dfrac{x-6}{x^2+1}.\dfrac{3x^2-3x+3}{x^2-36}+\dfrac{x-6}{x^2+1}.\dfrac{3x}{x^2-36}\)
\(=\dfrac{x-6}{x^2+1}\left(\dfrac{3x^2-3x+3}{x^2+1}+\dfrac{3x}{x^2+1}\right)\)
\(=\dfrac{x-6}{x^2+1}.\dfrac{3x^2+3}{x^2-36}\)
\(=\dfrac{x-6}{x^2+1}.\dfrac{3\left(x^2+1\right)}{\left(x-6\right)\left(x+6\right)}=\dfrac{3}{x+6}.\)
Tại \(x=994\) ta có \(A=\dfrac{3}{994+6}=\dfrac{3}{1000}.\)