ĐIỀU KIỆN XÁC ĐỊNH: \(x\ne1\)
\(S=\left(x-3+\dfrac{1}{x-1}\right):\left(x-1-\dfrac{1}{x-1}\right)\)
\(S=\left(\dfrac{x^2-x}{x-1}-\dfrac{3x-3}{x-1}+\dfrac{1}{x-1}\right):\left(\dfrac{x^2-x}{x-1}-\dfrac{x-1}{x-1}-\dfrac{1}{x-1}\right)\)\(S=\left(\dfrac{x^2-x-3x+3+1}{x-1}\right):\left(\dfrac{x^2-x-x+1-1}{x-1}\right)\)\(S=\left(\dfrac{x^2-4x+4}{x-1}\right):\left(\dfrac{x^2-2x}{x-1}\right)\)
\(S=\dfrac{\left(x-2\right)^2}{x-1}.\dfrac{x-1}{x\left(x-2\right)}\)
\(S=\dfrac{\left(x-2\right)^2.\left(x-1\right)}{\left(x-1\right).x.\left(x-2\right)}\)
\(S=\dfrac{x-2}{x}\)
b) ĐỂ S > 5 => \(\dfrac{x-2}{x}>5\)
\(\dfrac{x-2}{x}>\dfrac{5x}{x}\)
\(x-2>5x\)
\(x-5x>2\)
\(-4x>2\)
\(x< -\dfrac{1}{2}\)
VẬY ĐỂ S > 5 THÌ x < \(-\dfrac{1}{2}\)