\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x}=\dfrac{x^2+2x+6}{x\left(x-3\right)}\) đkxđ: x khác 3 , x khác 0
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-3\right)}+\dfrac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}-\dfrac{x^2+2x+6}{x\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+2x}{....}+\dfrac{x^2-3x-2x+6}{.....}-\dfrac{x^2+2x+6}{...}=0\)
\(\Leftrightarrow x^2+2x+x^2-3x-2x+6-x^2-2x-6=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
đk : x khác -3 ; 3
\(\Rightarrow\left(x+3\right)^2-4x^2=\left(x-3\right)^2\Leftrightarrow-3x^2+6x+9=x^2-6x+9\)
\(\Leftrightarrow4x^2-12x=0\Leftrightarrow4x\left(x-3\right)=0\Leftrightarrow x=3\left(ktm\right);x=0\)
\(\dfrac{x+3}{x-3}-\dfrac{4x^2}{x^2-9}=\dfrac{x-3}{x+3}\) đkxđ: x khác 3 ; x khác -3
\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+6x+9}{...}-\dfrac{4x^2}{.....}-\dfrac{x^2-6x+9}{....}=0\)
\(\Leftrightarrow x^2+6x+9-4x^2-x^2+6x-9=0\)
\(\Leftrightarrow-4x^2+12x=0\)
\(\Leftrightarrow-4x\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=3\left(kotm\right)\end{matrix}\right.\)
Vậy tập nghiệm của pt có S thuộc \(\left\{0\right\}\)
