\(\dfrac{3}{x+5}+\dfrac{x}{x-5}+\dfrac{x^2+5}{x^2-25}=0\left(x\ne\pm5\right)\\ \Leftrightarrow\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x^2+5}{\left(x+5\right)\left(x-5\right)}=0\\ \Leftrightarrow3\left(x-5\right)+x\left(x+5\right)=x^2+5\\ \Leftrightarrow3x-15+x^2+5x+x^2+5=0\\ \Leftrightarrow2x^2+8x-10=0\\ \Leftrightarrow2x^2-2x+10x-10=0\\ \Leftrightarrow2x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(2x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=-10\\x=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy: ...
ĐKXĐ: \(x\ne\pm5\)
\(\frac{3}{x+5}+\frac{x}{x-5}+\frac{x^2+5}{x^2-25}=0\\\Leftrightarrow \frac{3(x-5)}{(x-5)(x+5)}+\frac{x(x+5)}{(x-5)(x+5)}+\frac{x^2+5}{(x-5)(x+5)}=0\\\Rightarrow 3(x-5)+x(x+5)+x^2+5=0\\\Leftrightarrow 2x^2+8x-10=0\\\Leftrightarrow x^2+4x-5=0\\\Leftrightarrow x^2-x+5x-5=0\\\Leftrightarrow x(x-1)+5(x-1)=0\\\Leftrightarrow (x-1)(x+5)=0\\\Leftrightarrow \left[\begin{array}{} x+5=0\\ x-1=0 \end{array} \right. \Leftrightarrow\left[\begin{array}{} x=-5(ktm)\\ x=1(tm) \end{array}\right.\)
Vậy phương trình đã cho có 1 nghiệm duy nhất là x=1.
đk x khác -5;5
\(\dfrac{3}{x+5}+\dfrac{x}{x-5}+\dfrac{x^2+5}{x^2-25}=0\)
\(\Rightarrow3\left(x-5\right)+x\left(x+5\right)+x^2+5=0\Leftrightarrow x^2+8x-15+x^2+5=0\)
\(\Leftrightarrow2x^2+8x-10=0\Leftrightarrow x=1\left(tm\right);x=-5\left(l\right)\)
