a) ĐKXĐ : \(\hept{\begin{cases}x-3\ne0\\x^2-9\ne0\\x+3\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne3\\x\ne\pm3\\x\ne-3\end{cases}}\Rightarrow x\ne\pm3\)
b) A = \(\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}=\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)
Khi x = 3 => Không thỏa mãn ĐKXĐ
=> Không tồn tại A khi x = 3
a, Điều kiện xác định là :
\(\hept{\begin{cases}x-3\ne0\\x^2-9\ne0\\x+3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne3\\\left(x-3\right)\left(x+3\right)\ne\\x\ne-3\end{cases}}0\Rightarrow x\ne\pm3}\)
Vậy \(x\ne\pm3\)
b, \(A=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
\(=\frac{3}{x-3}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\)
\(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)
Thay x = 3 ( ktm đkxđ )
Ko tồn tại x
