\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x^2-3x-2x+6}+\frac{1}{x^2-3x-4x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-5x-6x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{1-5}-\frac{1}{1-4}+\frac{1}{1-4}-\frac{1}{1-3}+\frac{1}{1-3}-\frac{1}{1-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{x^2-8x+12}=\frac{1}{8}\)
\(\Leftrightarrow x^2-8x+12=32\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow x=10\) hoặc \(x=-2\)
\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)\(\frac{1}{x^2-2x-3x+6}+\frac{1}{x^2-4x-3x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-6x-5x+30}=\frac{1}{8}\)
\(\frac{1}{x\left(x-2\right)-3\left(x-2\right)}+\frac{1}{x\left(x-4\right)-3\left(x-4\right)}+\frac{1}{x\left(x-4\right)-5\left(x-4\right)}+\frac{1}{x\left(x-6\right)-5\left(x-6\right)}=\frac{1}{8}\)
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)dhjjhhjhhjj
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
Còn lại tự giải quyết nha