Question

Cho \(\frac{x+1}{\left(x+2\right)\left(x+3\right)}=\frac{a}{x+2}+\frac{b}{x+3}\) thì giá trị của b là bao nhiêu?

Answer 1

chi tiết:

\(\frac{a}{x+2}+\frac{b}{x+3}=\frac{a\left(x+3\right)+b\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}=\frac{ax+3a+bx+2b}{\left(x+2\right)\left(x+3\right)}=\frac{\left(a+b\right)x+3a+2b}{\left(x+2\right)\left(x+3\right)}\)

\(\frac{x+1}{\left(x+2\right)\left(x+3\right)}=\frac{\left(a+b\right)x+3a+2b}{\left(x+2\right)\left(x+3\right)}\Rightarrow\left\{\begin{matrix}a+b=1\\3a+2b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}3a+3b=3\\3a+2b=1\end{matrix}\right.\) trừ cho nhau => b=2; a=-1

Answer 2

b=2