Question

Answer 1

`a)`\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right).\left(1+\dfrac{1}{\sqrt{x}}\right)\)

\(ĐK:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

`b)`\(B=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(B=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(B=\dfrac{4}{\sqrt{x}-1}\)

\(B=1\)

\(\Leftrightarrow\dfrac{4}{\sqrt{x}-1}=1\)

\(\Leftrightarrow\sqrt{x}-1=4\)

`<=>x=25(tm)`