Bài 7:
ĐKXĐ: \(x\notin\left\{0;6;-6;3\right\}\)
a) Ta có: \(S=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)
\(=\left(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)
\(=\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)
\(=\dfrac{12x-36}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}-\dfrac{x}{x-6}\)
\(=\dfrac{12\left(x-3\right)}{2\left(x-3\right)\left(x-6\right)}-\dfrac{x}{x-6}\)
\(=\dfrac{6}{x-6}-\dfrac{x}{x-6}\)
\(=\dfrac{-\left(x-6\right)}{x-6}=1\)
b) Vì S=1 với mọi x thỏa mãn ĐKXĐ nên Không có giá trị nào của x để S=-1
