Câu hỏi của nasa

\(\dfrac{x^2+5}{x^3-3x-2}=\dfrac{a}{x+2}+\dfrac{b}{\left(x+1\right)^2}\left(1\right)\)Ta có :\(\dfrac{a}{x+2}+\dfrac{b}{\left(x+1\right)^2}\)\(=\dfrac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}\)\(=\dfrac{a\left(x^2+2x+1\right)+bx-2b}{\left(x^2-x-2\right)\left(x+1\right)}\)\(=\dfrac{ax^2+2ax+a+bx-2b}{x^3+x^2-x^2-x-2x-2}\)\(=\dfrac{ax^2+\left(2a+b\right)x+a-2b}{x^3-3x-2}\)\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}a=1\\2a+b=0\\a-2b=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\\1-2.\left(-2\right)=5\left(đúng\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\) thỏa mãn yêu cầu đề bàiCâu trả lời: \(\dfrac{x^2+5}{x^3-3x-2}=\dfrac{a}{x+2}+\dfrac{b}{\left(x+1\right)^2}\left(1\right)\)Ta có :\(\dfrac{a}{x+2}+\dfrac{b}{\left(x+1\right)^2}\)\(=\dfrac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}\)\(=\dfrac{a\left(x^2+2x+1\right)+bx-2b}{\left(x^2-x-2\right)\left(x+1\right)}\)\(=\dfrac{ax^2+2ax+a+bx-2b}{x^3+x^2-x^2-x-2x-2}\)\(=\dfrac{ax^2+\left(2a+b\right)x+a-2b}{x^3-3x-2}\)\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}a=1\\2a+b=0\\a-2b=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\\1-2.\left(-2\right)=5\left(đúng\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\) thỏa mãn yêu cầu đề bài