Chào tiểu thư ^_^ !
1)Rút gọn biểu thức
\(Q=\dfrac{1}{1^4+1^2+1}+\dfrac{2}{2^4+2^2+1}+...+\dfrac{n}{n^4+n^2+1}\)
\(Q=\dfrac{1}{1^4\left(1+1^{-2}+1^{-4}\right)}+\dfrac{2}{2^4\left(1+1^{-2}+1^{-4}\right)}+...+\dfrac{n}{n^4\left(1+1^{-2}+1^{-4}\right)}\\
\Leftrightarrow Q=\dfrac{1}{3}+\dfrac{1}{2^3.3}+...+\dfrac{1}{n^3.3}\\
\Leftrightarrow Q=\dfrac{1}{3}\left(\dfrac{1}{2^3}+...+\dfrac{1}{n^3}\right)\\
\Leftrightarrow Q=\dfrac{1}{3}\left(\dfrac{1}{8}+...+\dfrac{1}{n^3}\right)\)
2 Giải pt
\(\dfrac{4x^2+14}{x^2+6}-\dfrac{5}{x^2+1}=\dfrac{7}{x^2+3}+\dfrac{9}{x^2+5}\\
\Leftrightarrow\dfrac{4x^2+14}{x^2+6}-3-\dfrac{5}{x^2+1}+1-\dfrac{7}{x^2+3}+1-\dfrac{9}{x^2+5}+1=0\\
\Leftrightarrow\dfrac{4x^2+14-3x^2-18}{x^2+6}-\dfrac{5+x^2+1}{x^2+1}-\dfrac{7+x^2+3}{x^2+3}-\dfrac{9+x^2+5}{x^2+5}=0\\
\Leftrightarrow\dfrac{x^2-4}{x^2+6}-\dfrac{x^2-4}{x^2+1}-\dfrac{x^2-4}{x^2+3}-\dfrac{x^2-4}{x^2+5}=0\\
\Leftrightarrow\left(x^2-4\right)\left(\dfrac{1}{x^2+6}-\dfrac{1}{x^2+1}-\dfrac{1}{x^2+3}-\dfrac{1}{x^2+5}\right)=0\\
\Leftrightarrow x^2-4=0\\
\Leftrightarrow x^2=4\\
\Leftrightarrow x=2\)Vậy pt có nghiệm là x=2
Chúc tiểu thư học tốt ! 
TDVN2005.
