Ta có :
\(P=\left(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\right):\dfrac{1}{x^2-2x-8}\)
\(P=\left(\dfrac{8+x-4}{\left(x+4\right)\left(x-4\right)}\right):\dfrac{1}{\left(x+2\right)\left(x-4\right)}\)
\(P=\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}:\dfrac{1}{\left(x+2\right)\left(x-4\right)}\)
\(P=\dfrac{1}{x-4}.\left(x+2\right)\left(x-4\right)\)
\(P=\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x-4\right)}\)
\(P=x+2\)
2 . Ta có :
\(x^2-9x+20=0\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Thay \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) vào biểu thức \(P=x+2\) ta được :
\(\left[{}\begin{matrix}4+2=6\\5+2=7\end{matrix}\right.\)
Kết luận __________________________________
ĐKXĐ của phân thức là : \(\left\{{}\begin{matrix}x^2-16\ne0\\x+4\ne0\\x^2-2x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+4\right)\ne0\\x\ne-4\\\left(x-4\right)\left(x+2\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne-4\\x\ne-2\end{matrix}\right.\)
\(P=\left(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\right):\dfrac{1}{x^2-2x-8}\) \(=\left(\dfrac{8}{\left(x-4\right)\left(x+4\right)}+\dfrac{1}{x+4}\right).\left(x^2-2x-8\right)\) \(=\dfrac{8+x-4}{\left(x-4\right)\left(x+4\right)}.\left(x^2-4x+2x-8\right)\) \(=\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}.\left(x-4\right)\left(x+2\right)\) \(=x+2\) + Tính giá trị của P tại x2 - 9x + 20 = 0 \(x^2-9x+20=0\) \(\Rightarrow x^2-4x-5x+20=0\) \(\Rightarrow\left(x^2-4x\right)-\left(5x-20\right)=0\) \(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\) \(\Rightarrow\left(x-4\right)\left(x-5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\left(kot/m\right)\\x=5\left(t/m\right)\end{matrix}\right.\) Thay x = 5 vào biểu thức P ,có : \(5+2=7\) Vậy tại x= 5 giá trị của P là 7
