Đáp án:
\(\begin{array}{l}
6)\dfrac{{ - 2\sqrt y }}{{2\sqrt y - 1}}\\
7)\dfrac{1}{{\sqrt a + 2}}\\
8)\dfrac{{ - 4\sqrt y }}{{\sqrt y - 2}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)DK:x \ge 0;y \ne 4\\
F = \dfrac{{4\sqrt y \left( {\sqrt y - 2} \right) - 8y}}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}:\dfrac{{\sqrt y + 2 + 3\sqrt y - 6}}{{\sqrt y - 2}}\\
= \dfrac{{4y - 8\sqrt y - 8y}}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}.\dfrac{{\sqrt y - 2}}{{4\sqrt y - 2}}\\
= \dfrac{{ - 4y - 8\sqrt y }}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}.\dfrac{{\sqrt y - 2}}{{4\sqrt y - 2}}\\
= \dfrac{{ - 4\sqrt y \left( {\sqrt y + 2} \right)}}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}.\dfrac{{\sqrt y - 2}}{{2\left( {2\sqrt y - 1} \right)}}\\
= \dfrac{{ - 2\sqrt y }}{{2\sqrt y - 1}}\\
7)DK:a > 0\\
G = \dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a + 2} \right)}^2}}}:\dfrac{{a + a\sqrt a }}{{\sqrt a \left( {\sqrt a + 2} \right)}}\\
= \dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a + 2} \right)}^2}}}:\dfrac{{a + \sqrt a }}{{\sqrt a + 2}}\\
= \dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a + 2} \right)}^2}}}.\dfrac{{\sqrt a + 2}}{{\sqrt a \left( {\sqrt a + 1} \right)}}\\
= \dfrac{1}{{\sqrt a + 2}}\\
8)DK:y \ge 0;y \ne 4\\
H = \dfrac{{4\sqrt y \left( {\sqrt y - 2} \right) - 8y}}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}\\
= \dfrac{{4y - 8\sqrt y - 8y}}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}\\
= \dfrac{{ - 4y - 8\sqrt y }}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}\\
= \dfrac{{ - 4\sqrt y \left( {\sqrt y + 2} \right)}}{{\left( {\sqrt y - 2} \right)\left( {\sqrt y + 2} \right)}}\\
= \dfrac{{ - 4\sqrt y }}{{\sqrt y - 2}}
\end{array}\)
