\(\Leftrightarrow\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{5}\)
=>x(x+4)=5
\(\Leftrightarrow x^2+4x-5=0\)
=>(x+5)(x-1)=0
=>x=1 hoặc x=-5
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