Đáp án:
$\begin{array}{l}
A = {x^2} - \sqrt x + x - 1\\
= \sqrt x .\left( {x\sqrt x - 1} \right) + \left( {x - 1} \right)\\
= \sqrt x .\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right) + \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\\
= \left( {\sqrt x - 1} \right)\left( {\sqrt x .\left( {x + \sqrt x + 1} \right) + \sqrt x + 1} \right)\\
= \left( {\sqrt x - 1} \right)\left( {x\sqrt x + x + 2\sqrt x + 1} \right)\\
b)B = \sqrt {30} - \sqrt {29} \\
= \dfrac{{\left( {\sqrt {30} + \sqrt {29} } \right)\left( {\sqrt {30} - \sqrt {29} } \right)}}{{\sqrt {30} + \sqrt {29} }}\\
= \dfrac{{30 - 29}}{{\sqrt {30} + \sqrt {29} }} = \dfrac{1}{{\sqrt {30} + \sqrt {29} }}\\
C = \sqrt {29} - \sqrt {28} \\
= \dfrac{{29 - 28}}{{\sqrt {29} + \sqrt {28} }} = \dfrac{1}{{\sqrt {29} + \sqrt {28} }}\\
Do:\sqrt {30} + \sqrt {29} > \sqrt {29} + \sqrt {28} \\
\Rightarrow \dfrac{1}{{\sqrt {30} + \sqrt {29} }} < \dfrac{1}{{\sqrt {29} + \sqrt {28} }}\\
\Rightarrow B < C\\
Hay\,\sqrt {30} - \sqrt {29} < \sqrt {29} - \sqrt {28}
\end{array}$
