điều kiện \(\left(x>0;x\ne1\right)\)
a) B = \(\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
B = \(\left(\dfrac{\left(\sqrt{x}-1\right)+\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
B = \(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\) = \(\dfrac{2}{\sqrt{x}+1}\)
b) ta có : B \(\ge\) \(\dfrac{1}{2}\) \(\Leftrightarrow\) \(\dfrac{2}{\sqrt{x}+1}\) \(\ge\) \(\dfrac{1}{2}\) \(\Leftrightarrow\) \(\dfrac{2}{\sqrt{x}+1}\ge\dfrac{2}{4}\)
\(\Leftrightarrow\) \(\sqrt{x}+1\le4\) \(\Leftrightarrow\) \(\sqrt{x}\le3\) \(\Leftrightarrow\) \(x\le9\)
mà : \(x>0;x\ne1;và\in nguyêndương\)
vậy giá trị nguyên dương nào của x để B \(\ge\dfrac{1}{2}\) là : 9 ; 8 ; 7 ; 6 ; 5 ; 4 ; 3 ; 2 .
