Đáp án:
e) \(\left[ \begin{array}{l}
x = 49\\
x = 9
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)H = \left[ {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right].\dfrac{{2{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \left[ {\dfrac{{x + \sqrt x + 1 - x + \sqrt x - 1}}{{\sqrt x }}} \right].\dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x }}.\dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{4\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\
b)Thay:x = 8 + 2\sqrt 7 = 7 + 2\sqrt 7 .1 + 1\\
= {\left( {\sqrt 7 + 1} \right)^2}\\
\to H = \dfrac{{4\left( {\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} - 1} \right)}}{{\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} + 1}}\\
= \dfrac{{4\left( {\sqrt 7 + 1 - 1} \right)}}{{\sqrt 7 + 1 + 1}}\\
= \dfrac{{4\sqrt 7 }}{{\sqrt 7 + 2}} = \dfrac{{28 - 8\sqrt 7 }}{3}\\
c)H = \dfrac{7}{5}\\
\to \dfrac{{4\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}} = \dfrac{7}{5}\\
\to 10\left( {\sqrt x - 1} \right) = 7\sqrt x + 7\\
\to 10\sqrt x - 10 = 7\sqrt x + 7\\
\to 3\sqrt x = 17\\
\to \sqrt x = \dfrac{{17}}{3}\\
\to x = \dfrac{{289}}{9}\\
d)H > \dfrac{9}{7}\\
\to \dfrac{{4\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}} > \dfrac{9}{7}\\
\to \dfrac{{28\left( {\sqrt x - 1} \right) - 9\sqrt x - 9}}{{7\left( {\sqrt x + 1} \right)}} > 0\\
\to \dfrac{{19\sqrt x - 37}}{{7\left( {\sqrt x + 1} \right)}} > 0\\
\to 19\sqrt x - 37 > 0\left( {do:\sqrt x + 1 > 0\forall x > 0} \right)\\
\to \sqrt x > \dfrac{{37}}{{19}}\\
\to x > \dfrac{{1369}}{{361}}\\
e)H = \dfrac{{4\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}} = \dfrac{{4\left( {\sqrt x + 1} \right) - 8}}{{\sqrt x + 1}}\\
= 4 - \dfrac{8}{{\sqrt x + 1}}\\
H \in Z\\
\Leftrightarrow \dfrac{8}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 8\\
\sqrt x + 1 = 4\\
\sqrt x + 1 = 2\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 7\\
\sqrt x = 3\\
\sqrt x = 1\\
\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 49\\
x = 9\\
x = 1\left( l \right)\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
