4.\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{1}{9}\)
\(\Leftrightarrow2\left[\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}\right]=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+6}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{6}{x\left(x+6\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2x\left(x+6\right)=54\)
\(\Leftrightarrow2x^2+12x-54=0\)
\(\Leftrightarrow2x^2-6x+18x-54=0\)
\(\Leftrightarrow2x\left(x-3\right)+18\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+18\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x+9\right)=0\)
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 9 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -9
Vậy x = 3 hoặc x = -9.
