Đáp án:
B7:
b) \(0 < x \le \dfrac{{3 + \sqrt 5 }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
B6:\\
a)A = \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}} - \sqrt {5 - 2\sqrt 5 + 1} \\
= \sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 - \sqrt 5 + 1 = 1\\
B = \left[ {\dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right].\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{x - 1}}{x}\\
b)B < A\\
\to \dfrac{{x - 1}}{x} < 1\\
\to \dfrac{{x - 1 - x}}{x} < 0\\
\to \dfrac{{ - 1}}{x} < 0\\
\to x > 0;x \ne 1\\
B7:\\
a)A = \dfrac{{\sqrt 5 \left( {\sqrt 5 - 1} \right)}}{{\sqrt 5 - 1}} = \sqrt 5 \\
B = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }} + \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x + 2}}\\
= \sqrt x + 1 + \sqrt x - 2 = 2\sqrt x - 1\\
b)B \le A \to 2\sqrt x - 1 \le \sqrt 5 \\
\to \sqrt x \le \dfrac{{1 + \sqrt 5 }}{2}\\
\to 0 < x \le \dfrac{{3 + \sqrt 5 }}{2}
\end{array}\)
