a, \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\left(đkxđ:x>0,x\ne1\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(A=\left(\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(A=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b, A=1
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}=1\)
\(\Rightarrow\sqrt{x}+1=\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}-\sqrt{x}+1=0\)
\(\Leftrightarrow1=0\) ( vô lý)
vậy không có giá trị của x để A=1
