\(\dfrac{1}{x}+\dfrac{3}{x^2-3x}=\dfrac{x+3}{3-x}\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{3}{x\left(x-3\right)}=\dfrac{-x-3}{x-3}\) (1)
Đkxđ: \(x\ne0\) và \(x\ne3\)
(1)\(\Leftrightarrow\dfrac{x-3}{x\left(x-3\right)}+\dfrac{3}{x\left(x-3\right)}=\dfrac{x\left(-x-3\right)}{x\left(x-3\right)}\)
\(\Rightarrow x-3+3=-x^2-3x\)
\(\Leftrightarrow x^2+x+3x+3-3=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(loai\right)\\x=-4\left(nhan\right)\end{matrix}\right.\)
Vậy S={-4}
\(\dfrac{1}{x}\)+ \(\dfrac{3}{x^2-3x}\) = \(\dfrac{x+3}{3-x}\)
x(x-3) \(\ne\) 0 \(\Rightarrow\) x \(\ne\) 0 và x-3 \(\ne\) 0
\(\Rightarrow\) x \(\ne\) 0 và x \(\ne\) 3
ĐKXĐ: x\(\ne\) 0 ; x \(\ne\) 3
Phương trình \(\Leftrightarrow\) \(\dfrac{1}{x}+\dfrac{3}{x\left(x-3\right)}=\dfrac{-x-3}{x-3}\)
\(\Leftrightarrow\) \(\dfrac{x-3}{x\left(x-3\right)}\) + \(\dfrac{3}{x\left(x-3\right)}\) = \(\dfrac{x\left(-x-3\right)}{x\left(x-3\right)}\)
\(\Leftrightarrow\) x - 3 +3 = -x^2 -3x
\(\Leftrightarrow\) x^2 +4x =0
\(\Leftrightarrow\) x(x+4) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\left(loại\right)\\x=-4\left(thoảman\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S = \(\left\{-4\right\}\)