Giải thích các bước giải:
Ta có:
a,
B có nghĩa khi: \(\left\{ \begin{array}{l}
x \ge 0\\
x + \sqrt x + 1 \ne 0\\
\sqrt x \ne 0\\
\sqrt x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\)
b,
\(\begin{array}{l}
B = \dfrac{{{x^2} - \sqrt x }}{{x + \sqrt x + 1}} - \dfrac{{2x + \sqrt x }}{{\sqrt x }} + \dfrac{{2.\left( {x - 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x .\left( {x\sqrt x - 1} \right)}}{{x + \sqrt x + 1}} - \dfrac{{\sqrt x .\left( {2\sqrt x + 1} \right)}}{{\sqrt x }} + \dfrac{{2.\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x .\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}} - \left( {2\sqrt x + 1} \right) + 2.\left( {\sqrt x + 1} \right)\\
= \sqrt x .\left( {\sqrt x + 1} \right) - 2\sqrt x - 1 + 2\sqrt x + 2\\
= x + \sqrt x + 1\\
d,\\
\dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} - \sqrt {1 + \dfrac{{\sqrt 5 }}{3}} }} + \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} + \sqrt {1 - \dfrac{{\sqrt 5 }}{3}} }}\\
= \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} - \sqrt {\dfrac{{3 + \sqrt 5 }}{3}} }} + \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} + \sqrt {\dfrac{{3 - \sqrt 5 }}{3}} }}\\
= \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} - \sqrt {\dfrac{{6 + 2\sqrt 5 }}{6}} }} + \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} + \sqrt {\dfrac{{6 - 2\sqrt 5 }}{6}} }}\\
= \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} - \sqrt {\dfrac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{6}} }} + \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} + \sqrt {\dfrac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{6}} }}\\
= \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} - \dfrac{{\sqrt 5 + 1}}{{\sqrt 6 }}}} + \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{1}{{\sqrt 6 }} + \dfrac{{\sqrt 5 - 1}}{{\sqrt 6 }}}}\\
= \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{ - \dfrac{{\sqrt 5 }}{6}}} + \dfrac{{\dfrac{{\sqrt 5 }}{3}}}{{\dfrac{{\sqrt 5 }}{6}}}\\
= - 2 + 2 = 0
\end{array}\)
