Đáp án:
\(\begin{array}{l}
B5:\\
a)A = 1\\
B = \sqrt x - 1\\
b)0 < x < 4\\
B6:\\
a)A = 2\\
B = \dfrac{{2\sqrt x }}{3}\\
b)0 < x < 9;x \ne 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a)A = \dfrac{{3\left( {\sqrt 7 - 2} \right)}}{{7 - 4}} + \left| {\sqrt 7 - 3} \right|\\
= \sqrt 7 - 2 + 3 - \sqrt 7 = 1\\
B = \dfrac{{x - \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \sqrt x - 1\\
b)B < A\\
\to \sqrt x - 1 < 1\\
\to \sqrt x < 2\\
\to 0 < x < 4\\
B6:\\
a)A = \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - \sqrt {3 - 2\sqrt 3 .1 + 1} \\
= \sqrt 3 + 1 - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \sqrt 3 + 1 - \left( {\sqrt 3 - 1} \right)\\
= 2\\
B = \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) + \sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{3\sqrt x }}\\
= \dfrac{{x + 2\sqrt x + x - 2\sqrt x }}{{3\sqrt x }}\\
= \dfrac{{2x}}{{3\sqrt x }}\\
= \dfrac{{2\sqrt x }}{3}\\
b)A > B\\
\to 2 > \dfrac{{2\sqrt x }}{3}\\
\to \dfrac{{2\sqrt x - 6}}{3} < 0\\
\to 2\sqrt x - 6 < 0\\
\to \sqrt x < 3\\
\to 0 < x < 9;x \ne 4
\end{array}\)
