ĐKXĐ:\(x\ne3,4,7,12\)
\(\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{3}{\left(x-4\right)\left(x-7\right)}+\dfrac{5}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{x-12-x+3}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{-9}{x^2-15x+36}=\dfrac{1}{4}\\ \Leftrightarrow x^2-15x+36=-36\\ \Leftrightarrow x^2-15x+72=0\)
\(\Leftrightarrow x^2-2.\dfrac{15}{2}x+\dfrac{225}{4}+\dfrac{63}{4}=0\\ \Leftrightarrow\left(x-\dfrac{15}{2}\right)^2+\dfrac{63}{4}=0\left(vô.lí\right)\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-12}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{x-12-x+3}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\)
\(\Leftrightarrow\left(x-3\right)\left(x-12\right)=-36\)
=>x(x-15)=0
=>x=0 hoặc x=15