\(a,P=\dfrac{3}{x+4}+\dfrac{x}{x-4}-\dfrac{3x-12}{x^2-16}\left(MTC:\left(x-4\right)\left(x+4\right)\right)\)
\(=\dfrac{3\left(x-4\right)+x\left(x+4\right)-3x+12}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{3x-12+x^2+4x-3x+12}{\left(x-4\right)\left(x+4\right)}\)
\(\dfrac{x^2+4x}{\left(x-4\right)\left(x+4\right)}=\dfrac{x\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x}{x-4}\left(đpcm\right)\)
\(b,Thay\cdot x=-2\cdot vào\cdot P,ta\cdotđược:\)
\(\Rightarrow\dfrac{-2}{-2-4}=\dfrac{1}{3}\)
Vậy Giá trị của P tại x = -2 là \(\dfrac{1}{3}\)
a: \(P=\dfrac{3x-12+x^2+4x-3x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{x^2+4x}{\left(x-4\right)\left(x+4\right)}=\dfrac{x}{x-4}\)
b: Thay x=-2 vào P, ta được:
\(P=\dfrac{-2}{-2-4}=\dfrac{-2}{-6}=\dfrac{1}{3}\)
