a) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne0\\\sqrt{x}+2\ne0\end{matrix}\right.\Leftrightarrow x>0\)
\(A=\dfrac{x+\sqrt{x}}{\sqrt{x}}+\dfrac{x-4}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\sqrt{x}+1+\sqrt{x}-2\)
\(=2\sqrt{x}-1\)
b) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+1\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(B=\left(5-\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(5+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left[5-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right].\left[5+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)
\(=\left(5-\sqrt{x}\right).\left(5+\sqrt{x}\right)\)
\(=25-x\)
\(đkxđ:x>0.\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}.\)
\(A=\sqrt{x}+1+\sqrt{x}-2=2\sqrt{x}-1.\)
\(đkxđ:x>0;x\ne1.\\ B=\left(5-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(5+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\\ B=\left(5-\sqrt{x}\right)\left(5+\sqrt{x}\right).\\ B=25-x.\)
