Bài tập ôn tập chương 1

Ta có \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\) Suy ra

$\begin{array}{l}P <  - \dfrac{1}{2} \Leftrightarrow  - \dfrac{3}{{\sqrt x  + 3}} <  - \dfrac{1}{2}\\ \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} > \dfrac{1}{2} \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} - \dfrac{1}{2} > 0\\ \Leftrightarrow \dfrac{{6 - \sqrt x  - 3}}{{2\left( {\sqrt x  + 3} \right)}} > 0\\ \Leftrightarrow 3 - \sqrt x  > 0\,\,\,\left( {do\,\,\,\sqrt x  + 3 > 0\,\,\forall x \ge 0;\,x \ne 9} \right)\\ \Leftrightarrow \sqrt x  < 3 \Leftrightarrow x < 9.\end{array}$

Kết hợp với ĐKXĐ ta được với \(0 \le x < 9\) thì $P <  - \dfrac{1}{2}$.

Ta có: $x = \dfrac{{3 - \sqrt 5 }}{2} = \dfrac{{6 - 2\sqrt 5 }}{4} = \dfrac{{{{\left( {\sqrt 5  - 1} \right)}^2}}}{4}$

\(\begin{array}{l} \Rightarrow \sqrt x  = \sqrt {\dfrac{{{{\left( {\sqrt 5  - 1} \right)}^2}}}{4}}  = \dfrac{{\left| {\sqrt 5  - 1} \right|}}{2} = \dfrac{{\sqrt 5  - 1}}{2}.\\ \Rightarrow P = \dfrac{{ - 3}}{{\dfrac{{\sqrt 5  - 1}}{2} + 3}} = \dfrac{{ - 3.2}}{{\sqrt 5  - 1 + 6}} = \dfrac{{ - 6}}{{\sqrt 5  + 5}} = \dfrac{{ - 6\left( {5 - \sqrt 5 } \right)}}{{{5^2} - 5}} = \dfrac{{6\sqrt 5  - 30}}{{20}} = \dfrac{{3\sqrt 5  - 15}}{{10}}.\end{array}\)

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x  - 3 \ne 0\\x - 9 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 9\end{array} \right..\)

$\begin{array}{l}P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\\,\,\,\,\, = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \sqrt x \left( {\sqrt x  + 3} \right) - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{2x - 6\sqrt x  + x + 3\sqrt x  - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{ - 3\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\\,\,\,\,\, = \dfrac{{ - 3\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 1} \right)}} = \dfrac{{ - 3}}{{\sqrt x  + 3}}.\end{array}$

Vậy \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\)

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x  - 3 \ne 0\\x - 9 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 9\end{array} \right..\)

$\begin{array}{l}P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\\,\,\,\,\, = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \sqrt x \left( {\sqrt x  + 3} \right) - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{2x - 6\sqrt x  + x + 3\sqrt x  - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{ - 3\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\\,\,\,\,\, = \dfrac{{ - 3\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 1} \right)}} = \dfrac{{ - 3}}{{\sqrt x  + 3}}.\end{array}$

Vậy \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\)

Ta có \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\) Nên

 \(\begin{array}{l}K = \dfrac{{y + 81}}{{y - 81}} \Rightarrow \dfrac{{x + 9}}{{x - 9}} = \dfrac{{y + 81}}{{y - 81}}\\ \Rightarrow  \left( {x + 9} \right)\left( {y - 81} \right) = \left( {y + 81} \right)\left( {x - 9} \right)\\ \Leftrightarrow xy + 9y - 81x - 9.81 = xy - 9y + 81x - 9.81\\ \Leftrightarrow 9y = 81x\\ \Leftrightarrow \dfrac{y}{x} = \dfrac{{81}}{9} = 9.\end{array}\)

Vậy nếu \(K = \dfrac{{y + 81}}{{y - 81}}\) thì \(\dfrac{y}{x}\) là số nguyên chia hết cho 3.

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6 \ne 0\\\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  - 3} \right) \ne 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt x  \ne 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\x \ne 9\end{array} \right..\)

\(\begin{array}{l}K = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6}}\\ = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt y  + 2} \right)}}\\ = \dfrac{{\left( {2\sqrt x  + 3\sqrt y } \right)\left( {\sqrt x  + 3} \right) - \left( {6 - \sqrt {xy} } \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + 6\sqrt x  + 3\sqrt {xy}  + 9\sqrt y  - \left( {6\sqrt x  - 18 - x\sqrt y  + 3\sqrt {xy} } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + x\sqrt y  + 9\sqrt y  + 18}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}} = \dfrac{{\left( {\sqrt y  + 2} \right)\left( {x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{x + 9}}{{x - 9}}.\end{array}\)

Vậy \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\)

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6 \ne 0\\\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  - 3} \right) \ne 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt x  \ne 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\x \ne 9\end{array} \right..\)

\(\begin{array}{l}K = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6}}\\ = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt y  + 2} \right)}}\\ = \dfrac{{\left( {2\sqrt x  + 3\sqrt y } \right)\left( {\sqrt x  + 3} \right) - \left( {6 - \sqrt {xy} } \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + 6\sqrt x  + 3\sqrt {xy}  + 9\sqrt y  - \left( {6\sqrt x  - 18 - x\sqrt y  + 3\sqrt {xy} } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + x\sqrt y  + 9\sqrt y  + 18}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}} = \dfrac{{\left( {\sqrt y  + 2} \right)\left( {x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{x + 9}}{{x - 9}}.\end{array}\)

Vậy \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\)

Vì $x = 25$ (TMĐK) nên ta có: $\sqrt x  = 5$

Khi đó ta có: $A = \dfrac{7}{{5 + 8}} = \dfrac{7}{{13}}$

Điều kiện \(x > 0,x \ne 4\)

$\begin{array}{l}A = \left( {\dfrac{1}{{\sqrt x }} - \dfrac{{\sqrt x  - 1}}{{x + 2\sqrt x }}} \right):\left( {\dfrac{1}{{\sqrt x  + 2}} - \dfrac{{\sqrt x  + 1}}{{x - 4}}} \right)\\ = \left( {\dfrac{{\sqrt x  + 2}}{{\sqrt x \left( {\sqrt x  + 2} \right)}} - \dfrac{{\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}} \right):\left( {\dfrac{{\sqrt x  - 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}} - \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}} \right)\\ = \dfrac{3}{{\sqrt x \left( {\sqrt x  + 2} \right)}}:\dfrac{{ - 3}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\ = \dfrac{3}{{\sqrt x \left( {\sqrt x  + 2} \right)}}.\dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}{{ - 3}}\\ = \dfrac{{2 - \sqrt x }}{{\sqrt x }}\end{array}$

Vậy với \(x > 0,x \ne 4\)  thì  \(A = \dfrac{{2 - \sqrt x }}{{\sqrt x }}\).

\(\begin{array}{l}P = \dfrac{{\sqrt {3 - \sqrt 5 } .\left( {3 + \sqrt 5 } \right)}}{{\sqrt {10}  + \sqrt 2 }} \\= \dfrac{{\sqrt {3 - \sqrt 5 } \left( {3 + \sqrt 5 } \right)\left( {\sqrt {10}  - \sqrt 2 } \right)}}{{10 - 2}}\\ = \dfrac{{\sqrt {3 - \sqrt 5 } \left( {3 + \sqrt 5 } \right).\sqrt 2 \left( {\sqrt 5  - 1} \right)}}{8}\\ = \dfrac{{\sqrt {6 - 2\sqrt 5 } .\left( {3\sqrt 5  + 5 - 3 - \sqrt 5 } \right)}}{8} \\= \dfrac{{\sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}} .\left( {2\sqrt 5  + 2} \right)}}{8}\\ = \dfrac{{\left( {\sqrt 5  - 1} \right).2.\left( {\sqrt 5  + 1} \right)}}{8} \\= \dfrac{{2.\left( {5 - 1} \right)}}{8} \\= 1.\,\,\,\left( {do\,\,\,\sqrt 5  - 1 > 0} \right).\end{array}\)

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x  - 3 \ne 0\\x - 9 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 9\end{array} \right..\)

$\begin{array}{l}P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\\,\,\,\,\, = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \sqrt x \left( {\sqrt x  + 3} \right) - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{2x - 6\sqrt x  + x + 3\sqrt x  - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{ - 3\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\\,\,\,\,\, = \dfrac{{ - 3\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 1} \right)}} = \dfrac{{ - 3}}{{\sqrt x  + 3}}.\end{array}$

Vậy \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\)

ĐKXĐ: \(x \ge 0,\,\,x \ne 4.\)

\(Q = \dfrac{{2x - 3\sqrt x  - 2}}{{\sqrt x  - 2}}\)\( = \dfrac{{\left( {2\sqrt x  + 1} \right)\left( {\sqrt x  - 2} \right)}}{{\sqrt x  - 2}}\)\( = 2\sqrt x  + 1.\)

Ta có: \(x = 2020 - 2\sqrt {2019} \)\( = 2019 - 2\sqrt {2019}  + 1\)\( = {\left( {\sqrt {2019}  - 1} \right)^2}\,\,\,\left( {tm} \right)\)

\( \Rightarrow \sqrt x  = \sqrt {{{\left( {\sqrt {2019}  - 1} \right)}^2}}  = \left| {\sqrt {2019}  - 1} \right| = \sqrt {2019}  - 1\,.\)

Thay \(\sqrt x  = \sqrt {2019}  - 1\) vào biểu thức \(Q\) ta được:

\(Q = 2\left( {\sqrt {2019}  - 1} \right) + 1\)\( = 2\sqrt {2019}  - 2 + 1\)\( = 2\sqrt {2019}  - 1.\)

Vậy \(x = 2020 - 2\sqrt {2019} \) thì \(Q = 2\sqrt {2019}  - 1.\)

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6 \ne 0\\\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  - 3} \right) \ne 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt x  \ne 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\x \ne 9\end{array} \right..\)

\(\begin{array}{l}K = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6}}\\ = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt y  + 2} \right)}}\\ = \dfrac{{\left( {2\sqrt x  + 3\sqrt y } \right)\left( {\sqrt x  + 3} \right) - \left( {6 - \sqrt {xy} } \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + 6\sqrt x  + 3\sqrt {xy}  + 9\sqrt y  - \left( {6\sqrt x  - 18 - x\sqrt y  + 3\sqrt {xy} } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + x\sqrt y  + 9\sqrt y  + 18}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}} = \dfrac{{\left( {\sqrt y  + 2} \right)\left( {x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{x + 9}}{{x - 9}}.\end{array}\)

Vậy \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\)

Với \(a > 0,\,\,a \ne 1\) ta có:
\(\begin{array}{l}B = \dfrac{{\sqrt a + 1}}{{a\sqrt a + a + \sqrt a }}:\dfrac{1}{{{a^2} - \sqrt a }}\\B = \dfrac{{\sqrt a + 1}}{{\sqrt a \left( {a + \sqrt a + 1} \right)}}.\sqrt a \left( {a\sqrt a - 1} \right)\\B = \dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{a + \sqrt a + 1}}\\B = \left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) = a - 1\end{array}\)

Điều kiện: \(x \ge 0;x \ne 4;x \ne 9.\)
\(\begin{array}{l}\left( {x - 9} \right).B < 2x \Leftrightarrow \left( {x - 9} \right).\dfrac{1}{{\sqrt x - 3}} < 2x\\ \Leftrightarrow \sqrt x + 3 < 2x \Leftrightarrow 2x - \sqrt x - 3 > 0\\ \Leftrightarrow \left( {\sqrt x + 1} \right)\left( {2\sqrt x - 3} \right) > 0\\ \Leftrightarrow 2\sqrt x - 3 > 0\,\,\,\left( {do\,\,\,\sqrt x + 1 > 0} \right)\\ \Leftrightarrow \sqrt x > \dfrac{3}{2} \Leftrightarrow x > \dfrac{9}{4}.\end{array}\)
Kết hợp điều kiện, ta được \(x > \dfrac{9}{4};x \ne 4;x \ne 9\) thỏa mãn yêu cầu đề bài.
Vậy \(x > \dfrac{9}{4},\,\,x \ne 4,\,\,x \ne 9\) thỏa mãn điều kiện bài toán.

Với \(x \ge 0;x \ne 4;x \ne 9,\) ta có:
\(\begin{array}{l}B = \dfrac{{5\sqrt x - 9}}{{x - 5\sqrt x + 6}} + \dfrac{{\sqrt x + 2}}{{3 - \sqrt x }} + \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}}\\B = \dfrac{{5\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} + \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}}\\B = \dfrac{{5\sqrt x - 9 - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\B = \dfrac{{5\sqrt x - 9 - x + 4 + x - 4\sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\B = \dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\B = \dfrac{1}{{\sqrt x - 3}}.\end{array}\)
Vậy \(B = \dfrac{1}{{\sqrt x - 3}}\) với \(x \ge 0;x \ne 4;x \ne 9.\)

Điều kiện xác định: \(x \ge 0.\)

Thay \(x = 25\,\,\,\left( {tm} \right)\) vào biểu thức ta có: \(A = \dfrac{{\sqrt {25} }}{{\sqrt {25}  + 1}} = \dfrac{5}{6}.\)

Vậy \(x = 25\) thì \(A = \dfrac{5}{6}.\)

Điều kiện xác định: \(x \ge 0.\)

Thay \(x = 25\,\,\,\left( {tm} \right)\) vào biểu thức ta có: \(A = \dfrac{{\sqrt {25} }}{{\sqrt {25}  + 1}} = \dfrac{5}{6}.\)

Vậy \(x = 25\) thì \(A = \dfrac{5}{6}.\)

Điều kiện: \(x \ge 0.\)
\(\begin{array}{l}P = \left( {\dfrac{{3\sqrt x }}{{x\sqrt x + 1}} - \dfrac{{\sqrt x }}{{x - \sqrt x + 1}} + \dfrac{1}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x + 3}}{{x - \sqrt x + 1}}\\\,\,\,\, = \left[ {\dfrac{{3\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} - \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} + \dfrac{{x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right]:\dfrac{{\sqrt x + 3}}{{x - \sqrt x + 1}}\\\,\,\,\, = \dfrac{{3\sqrt x - x - \sqrt x + x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{x - \sqrt x + 1}}{{\sqrt x + 3}}\\\,\,\,\, = \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{x - \sqrt x + 1}}{{\sqrt x + 3}} \\= \dfrac{1}{{\sqrt x + 3}}.\end{array}\)
\(\begin{array}{l} \Rightarrow P \ge \dfrac{1}{5} \Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \ge \dfrac{1}{5}\\ \Leftrightarrow \dfrac{1}{{\sqrt x + 3}} - \dfrac{1}{5} \ge 0 \Leftrightarrow \dfrac{{5 - \sqrt x - 3}}{{5\left( {\sqrt x + 3} \right)}} \ge 0\\ \Leftrightarrow \dfrac{{2 - \sqrt x }}{{5\left( {\sqrt x + 3} \right)}} \ge 0 \Leftrightarrow 2 - \sqrt x \ge 0\\ \Leftrightarrow \sqrt x \le 2 \Leftrightarrow x \le 4\end{array}\)
Vậy \(0 \le x \le 4\) thỏa mãn bài toán.

Điều kiện: \(a > 0,\,\,a \ne 4.\)
\(P = A.B = \dfrac{{a - 4}}{{a + 2\sqrt a }}.\dfrac{{a + 7\sqrt a }}{{a - 4}}\)\( = \dfrac{{\sqrt a \left( {\sqrt a + 7} \right)}}{{\sqrt a \left( {\sqrt a + 2} \right)}}\)
\( = \dfrac{{\sqrt a + 7}}{{\sqrt a + 2}} = \dfrac{{\sqrt a + 2 + 5}}{{\sqrt a + 2}}\)\( = 1 + \dfrac{5}{{\sqrt a + 2}} > 1\)
Ta có: với \(a > 0 \Rightarrow \sqrt a > 0 \Rightarrow \sqrt a + 2 > 2\)
\(\begin{array}{l} \Rightarrow \dfrac{1}{{\sqrt a + 2}} < \dfrac{1}{2} \Rightarrow \dfrac{5}{{\sqrt a + 2}} < \dfrac{5}{2}\\ \Rightarrow P = 1 + \dfrac{5}{{\sqrt a + 2}} < 1 + \dfrac{5}{2} = \dfrac{7}{2}\\ \Rightarrow 1 < P < \dfrac{7}{2}\end{array}\)
Mà \(P \in \mathbb{Z} \Rightarrow P = \left\{ {2;\,\,3} \right\}.\)
+) Với \(P = 2 \Leftrightarrow \dfrac{{\sqrt a + 7}}{{\sqrt a + 2}} = 2\) \( \Leftrightarrow \sqrt a + 7 = 2\left( {\sqrt a + 2} \right)\) \( \Leftrightarrow \sqrt a + 7 = 2\sqrt a + 4\)\( \Leftrightarrow \sqrt a = 3 \Leftrightarrow a = 9\,\,\,\left( {tm} \right).\)
+) Với \(P = 3 \Leftrightarrow \dfrac{{\sqrt a + 7}}{{\sqrt a + 2}} = 3\) \( \Leftrightarrow \sqrt a + 7 = 3\left( {\sqrt a + 2} \right)\)\( \Leftrightarrow \sqrt a + 7 = 3\sqrt a + 6\)\( \Leftrightarrow 2\sqrt a = 1 \Leftrightarrow \sqrt a = \dfrac{1}{2} \Leftrightarrow a = \dfrac{1}{4}\,\,\,\left( {tm} \right).\)
Vậy \(a = 9\) và \(a = \dfrac{1}{4}\) thỏa mãn yêu cầu bài toán.