Câu hỏi của Khánh Chi Trần

ĐKXĐ: \(x\notin\left\{1;-1;-\dfrac{1}{2}\right\}\)\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)\(=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)\(=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)\(=\dfrac{x^2-2x+1-x^2-x+3x+1}{2x+1}=\dfrac{2}{2x+1}\)Câu trả lời: ĐKXĐ: \(x\notin\left\{1;-1;-\dfrac{1}{2}\right\}\)\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)\(=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)\(=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)\(=\dfrac{x^2-2x+1-x^2-x+3x+1}{2x+1}=\dfrac{2}{2x+1}\)