a/
\(A=\dfrac{\sqrt{x}}{x+\sqrt{x}}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{x+\sqrt{x}}\right)\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)
= \(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}.\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{2\sqrt{x}}.\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
= \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}}\)
b/ Với \(x>0,x\ne1\)
Để A=1 \(\Leftrightarrow\dfrac{1}{\sqrt{x}}=1\)
\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(ktm\right)\)
Vậy không có giá trị nào của x thỏa mãn A=1
