\(ĐKXĐ:x>1\)
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(A=\dfrac{x-1}{2\sqrt{x}}.\left[\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)+\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\dfrac{x-1}{2\sqrt{x}}.\left(\dfrac{x\sqrt{x}-x-x+\sqrt{x}+x\sqrt{x}+x+x+\sqrt{x}}{x-1}\right)\)
\(A=\dfrac{x-1}{2\sqrt{x}}.\dfrac{2x\sqrt{x}+2\sqrt{x}}{x-1}\)
\(A=\dfrac{2x\sqrt{x}+2\sqrt{x}}{2\sqrt{x}}=\dfrac{2\sqrt{x}\left(x+1\right)}{2\sqrt{x}}=x+1\)
b. Để: \(A=x+1>-6\Rightarrow x+1+6>0\)
\(\Rightarrow x+7>0\Rightarrow x>-7\)
Mà: theo ĐKXĐ: \(x>1\)
⇒ \(x\in R\&x>1\)
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\\ ĐKXĐ:x>0;x\ne1\\ \Rightarrow A=\left(\dfrac{x}{2}-\dfrac{1}{2}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\\ =\dfrac{x-1}{2}\cdot\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\\ =\dfrac{x-1}{2}\cdot\dfrac{x-2\sqrt{x}+1+x+2\sqrt{x}+1}{x-1}\\ =\dfrac{2x+2}{2}=x+1\)
b) Với \(x>0;x\ne1\)
Để \(A>-6\)
thì \(\Rightarrow x+1>-6\)
\(\Rightarrow x>-7\)
Bài này dễ mà, bạn tự làm đi, đừng hỏi, hãy tự suy nghĩ sẽ giúp bạn giỏi hơn ^_^
