a, x khác 0 ; -3 ; 3 ; -1
b, \(P=\dfrac{2x\left(x+1\right)\left(x-3\right)^2}{x\left(x^2-9\right)\left(x+1\right)}=\dfrac{2\left(x-3\right)}{x+3}\)
Thay x = 1/2 vào P ta được
\(\dfrac{2\left(\dfrac{1}{2}-3\right)}{\dfrac{1}{2}+3}=-\dfrac{10}{7}\)
c, Ta có : P = 0 <=> \(\dfrac{2\left(x-3\right)}{x+3}=0\Rightarrow x-3=0\Leftrightarrow x=3\)(ktm)
Vậy ko có giá trị x thỏa mãn P = 0
a)P xác định \(< =>\left\{{}\begin{matrix}x\ne0\\x^2\ne9\\x+1\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne0\\x\ne\pm3\\x\ne-1\end{matrix}\right.\)
b)Với \(x\ne0,x\ne\pm3,x\ne-1\)
P=\(\dfrac{\left(2x^2+2x\right)\left(x-3\right)^2}{x\left(x^2-9\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-3\right)^2}{x\left(x+3\right)\left(x-3\right)\left(x+1\right)}=\dfrac{2\left(x-3\right)}{x+3}\)
Ta có:\(x=\dfrac{1}{2}\left(TM\right)\)
Thay \(x=\dfrac{1}{2}\) vào pt P đã rút gọn ta được:
P=\(\dfrac{2\left(\dfrac{1}{2}-3\right)}{\dfrac{1}{2}+3}=\dfrac{2.\dfrac{-5}{2}}{\dfrac{7}{2}}=-5.\dfrac{2}{7}=\dfrac{-10}{7}\)
c)Với \(x\ne0,x\ne\pm3,x\ne-1\)
P=0<=>\(\dfrac{2\left(x-3\right)}{x+3}=0\)
<=>2(x-3)=0(Vì \(x\ne-3\))
<=>x-3=0
<=>x=3(Không thỏa mãn)