Question

giải phương trình:\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{3}{10}\)

Answer 1

Ta có : \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{3}{10}\)

\(\Leftrightarrow10\left(x+3\right)-10x=3x\left(x+3\right)\)

\(\Leftrightarrow-3x^2-9x+30=0\)

\(\Delta=\left(-9\right)^2+4.3.30=81+360=441>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{9+\sqrt{441}}{-6}=-5\\x_2=\dfrac{9-\sqrt{441}}{-6}=2\end{matrix}\right.\)

Vậy \(S=\left\{-5;2\right\}\)