\(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\\ ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\\ \Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{52\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}-\dfrac{52\left(x+1\right)}{52\left(x+1\right)\left(x+10\right)}=\dfrac{9\left(x+1\right)\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}\\ \Leftrightarrow52\left(x+10\right)-52\left(x+1\right)=9\left(x+1\right)\left(x+10\right)\\ \Leftrightarrow9\left(x^2+10x+x+10\right)=52\left(x+10-x-1\right)\\ \Leftrightarrow9\left(x^2+11x+10\right)=52\cdot9\\ \Leftrightarrow x^2+11x+10=52\\ \Leftrightarrow x^2+14x-3x-42=0\\ \Leftrightarrow x\left(x+14\right)-3\left(x+14\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+14\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\left(T/m\right)\)
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