Ta có : \(\dfrac{x+1}{\left(x+2\right)\left(x+3\right)}=\dfrac{a}{x+2}+\dfrac{b}{x+3}\)
\(=\dfrac{a\left(x+3\right)+b\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{ax+3a+bx+2b}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x+1}{\left(x+2\right)\left(x+3\right)}=\dfrac{\left(a+b\right)x+3a+2b}{\left(x+2\right)\left(x+3\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=1\\3a+2b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=1-b\\3\left(1-b\right)+2b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\\3-3b+2b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\\3-b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=2\end{matrix}\right.\)
Vậy giá trị của b là 2
