Đáp án:
\(\begin{array}{l}
1) - \dfrac{1}{7}\\
2)\dfrac{{2{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
3)0 < x < 9;x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = \dfrac{9}{{16}}\\
\to Q = \dfrac{{\sqrt {\dfrac{9}{{16}}} - 1}}{{\sqrt {\dfrac{9}{{16}}} + 1}} = \dfrac{{\dfrac{3}{4} - 1}}{{\dfrac{3}{4} + 1}}\\
= - \dfrac{1}{7}\\
2)P = \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}} - \dfrac{4}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1 + x - \sqrt x + 1}}{{\sqrt x }} - \dfrac{4}{{\sqrt x }}\\
= \dfrac{{2x + 2 - 4}}{{\sqrt x }} = \dfrac{{2x - 2}}{{\sqrt x }}\\
A = P.Q = \dfrac{{2x - 2}}{{\sqrt x }}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{2{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
3)A.\sqrt x < 8\\
\to \dfrac{{2{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}.\sqrt x < 8\\
\to 2{\left( {\sqrt x - 1} \right)^2} < 8\\
\to {\left( {\sqrt x - 1} \right)^2} < 4\\
\to - 2 < \sqrt x - 1 < 2\\
\to - 1 < \sqrt x < 3\\
\to 0 < x < 9;x \ne 1
\end{array}\)
