Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A + B = \left( {9 + 3\sqrt 7 } \right) + \left( {9 - 3\sqrt 7 } \right) = 18\\
A.B = \left( {9 + 3\sqrt 7 } \right).\left( {9 - 3\sqrt 7 } \right) = {9^2} - {\left( {3\sqrt 7 } \right)^2} = {9^2} - 9.7 = 9.\left( {9 - 7} \right) = 9.2 = 18\\
\Rightarrow A + B = A.B\\
b,\\
y = \dfrac{{{x^2} + \sqrt x }}{{x - \sqrt x + 1}} - \dfrac{{2x + \sqrt x }}{{\sqrt x }} + 1\\
= \dfrac{{\sqrt x .\left( {x\sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - \dfrac{{\sqrt x \left( {2\sqrt x + 1} \right)}}{{\sqrt x }} + 1\\
= \dfrac{{\sqrt x .\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - \left( {2\sqrt x + 1} \right) + 1\\
= \sqrt x .\left( {\sqrt x + 1} \right) - \left( {2\sqrt x + 1} \right) + 1\\
= x + \sqrt x - 2\sqrt x - 1 + 1\\
= x - \sqrt x \\
2,\\
y = x - \sqrt x = \sqrt x .\left( {\sqrt x - 1} \right)\\
x > 1 \Rightarrow \sqrt x - 1 > 0 \Rightarrow y = \sqrt x \left( {\sqrt x - 1} \right) > 0\\
\Rightarrow \left| y \right| = y \Leftrightarrow y - \left| y \right| = 0
\end{array}\)
