Đáp án:
\(\begin{array}{l}
a)M = \sqrt 2 \\
N = \dfrac{{3x - 4}}{{x - 4}}\\
b)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)M = \dfrac{{7\left( {\sqrt 3 + \sqrt 2 } \right)}}{{3 - 2}} - 7\sqrt 3 - 2.3\sqrt 2 \\
= 7\sqrt 3 + 7\sqrt 2 - 7\sqrt 3 - 6\sqrt 2 \\
= \sqrt 2 \\
N = \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x - 2}} - \dfrac{{5\sqrt x + 2}}{{x - 4}}\\
= \dfrac{{\left( {3\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 5\sqrt x - 2 - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x - 4}}{{x - 4}}\\
b)N = {M^2}\\
\to \dfrac{{3x - 4}}{{x - 4}} = 2\\
\to 3x - 4 = 2x - 8\\
\to x = - 4\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)
