Ta có:
\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)
pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\) (tm)
