ê sẽ trả loi nhưng đổi lại cho a in4 nhé
\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x+3}=\dfrac{2\left(x^2+6\right)}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow x^2+2x+3x+6+x^2-2x-3x+6=2x^2+12\)
\(\Leftrightarrow2x^2+12=2x^2+12\) (luôn đúng)
-Vậy S={x∈R/x≠2;x≠-2}
\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x+3}=\dfrac{2\left(x^2+6\right)}{x^2-9}\) \(\left(đk:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+3\right)+\left(x-2\right)\left(x-3\right)-2\left(x^2+6\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+3x+2x+6+x^2-3x-2x+6-2x^2-12=0\)
\(\Leftrightarrow0=0\)
\(\Rightarrow x=0\left(n\right)\)
Vậy \(S=\left\{0\right\}\)