mình nghĩ đề sai .vậy mới đúng nè:
\(\dfrac{5+x}{5-x}=\dfrac{25-x^2}{x^2-10x+25}\)
<=> \(\dfrac{\left(x+5\right)}{\left(x-5\right)}-\dfrac{5^2-x^2}{x^2-2.5.x+5^2}=0\)
<=> \(\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\left(x-5\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x-5\right)^2}=0\) (đúng với mọi x)
\(\dfrac{x^3-9x}{18-6x}\) hay \(\dfrac{x^3-9x}{6x-18}\) vậy
\(\dfrac{x^3-9x}{18-6x}=\dfrac{x^2-3x}{6}\)
\(\Leftrightarrow\dfrac{x\left(x^2-9\right)}{6\left(3-x\right)}-\dfrac{x\left(x-3\right)}{6}=0\)
\(\Leftrightarrow\dfrac{-x\left(x-3\right)\left(x+3\right)}{6\left(x-3\right)}-\dfrac{x\left(x-3\right)}{6}=0\)
\(\Leftrightarrow-\dfrac{x\left(x+3\right)}{6}-\dfrac{x\left(x-3\right)}{6}=0\)
\(\Leftrightarrow\dfrac{-x^2-3-x^2+3}{6}=0\Leftrightarrow-2x^2=0\Leftrightarrow x=0\)
Vậy để hai phân thức trên bằng nhau thì \(x=0\)