Đáp án:
\(\begin{array}{l}
a)\dfrac{{11 - \sqrt 3 }}{{\sqrt {4 - \sqrt 3 } - 5}}\\
b)N = \dfrac{{\sqrt x }}{{\sqrt x - 5}}\\
c)Min = 2\sqrt 7
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = \sqrt {19 - 8\sqrt 3 } = \sqrt {16 - 2.4.\sqrt 3 + 3} \\
= \sqrt {{{\left( {4 - \sqrt 3 } \right)}^2}} = 4 - \sqrt 3 \\
\to M = \dfrac{{4 - \sqrt 3 + 7}}{{\sqrt {4 - \sqrt 3 } - 5}}\\
= \dfrac{{11 - \sqrt 3 }}{{\sqrt {4 - \sqrt 3 } - 5}}\\
b)N = \dfrac{{\sqrt x \left( {\sqrt x - 5} \right) + \left( {2\sqrt x - 3} \right)\left( {\sqrt x + 5} \right) - 2x + 3\sqrt x + 15}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{x - 5\sqrt x + 2x + 7\sqrt x - 15 - 2x + 3\sqrt x + 15}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{x + 5\sqrt x }}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x - 5}}\\
c)P = M:N = \dfrac{{x + 7}}{{\sqrt x - 5}}:\dfrac{{\sqrt x }}{{\sqrt x - 5}}\\
= \dfrac{{x + 7}}{{\sqrt x }} = \sqrt x + \dfrac{7}{{\sqrt x }}\\
Xét:x > 0\\
BDT:Co - si:\sqrt x + \dfrac{7}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{7}{{\sqrt x }}} \\
\to \sqrt x + \dfrac{7}{{\sqrt x }} \ge 2\sqrt 7 \\
\to Min = 2\sqrt 7 \\
\Leftrightarrow \sqrt x = \dfrac{7}{{\sqrt x }}\\
\to x = 7
\end{array}\)
