\(A=(\dfrac{2x+1}{x-1}+\dfrac{7}{x^2-1}-\dfrac{x-1}{x+1}).\dfrac{x^2-1}{2}\)
\(A=(\dfrac{2x+1}{x-1}+\dfrac{7}{(x-1)(x+1)}-\dfrac{x-1}{x+1}).\dfrac{(x-1)(x+1)}{2}\)
\(A=(\dfrac{2x^2+3x+1}{(x-1)(x+1)}+\dfrac{7}{(x-1)(x+1)}-\dfrac{x^2-2x+1}{(x-1)(x+1)}).\dfrac{(x-1)(x+1)}{2}\)
\(A=\dfrac{2x^2+3x+1+7-x^2+2x-1}{(x-1)(x+1)}.\dfrac{(x-1)(x+1)}{2}\)
\(A=\dfrac{x^2+5x+7}{(x-1)(x+1)}.\dfrac{(x-1)(x+1)}{2}\)
\(A=\dfrac{x^2+5x+7}{2}\)
Để \(A\in\mathbb{N*}, x\in\mathbb{R}, x\neq\pm 1\)
\(\Rightarrow A>0 x\in\mathbb{R}, x\neq\pm 1\)
\(\Rightarrow \dfrac{x^2+5x+7}{2}>0 x\neq\pm 1\)
\(\Rightarrow x^2+5x+7>0\)
\(\Rightarrow x^2+5x+7=x^2+5x+(\dfrac{5}{2})^2+\dfrac{3}{4}=[x^2+2.x.\dfrac{5}{2}+(\dfrac{5}{2})]+\dfrac{3}{4}=(x+\dfrac{5}{2})^2)+\dfrac{3}{4}\)
\(\Rightarrow (x+\dfrac{5}{2}))^2+\dfrac{3}{4}\geq \dfrac{3}{4}\forall x\in\mathbb{R}, x\neq\pm 1\)
\(\Rightarrow \dfrac{x^2+5x+7}{2}\geq\dfrac{3}{4}\forall x\in\mathbb{R}, x\neq\pm 1\)
Mà \(\dfrac{3}{4}>0\)
\(\Rightarrow\dfrac{x^2+5x+7}{2}>0\forall x\in\mathbb{R}, x\neq\pm 1\)
\(\Rightarrow A>0\forall x\in\mathbb{R}, x\neq\pm 1\)
\(\Rightarrow\) A luôn nhận giá trị dương.
