a)Cho biểu thức \(P = \left( {\frac{{3 - \sqrt x }}{{\sqrt x - 2}} + \frac{{\sqrt x - 2}}{{\sqrt x + 3}} - \frac{{9 - x}}{{x + \sqrt x - 6}}} \right)\) :\(\left( {\frac{{x - 3\sqrt x }}{{x - 9}} + \frac{{\sqrt x + 1}}{{x + 4\sqrt x + 3}}} \right)\) với \(x \ge 0,x \ne 4,x \ne 9\)

Tìm các giá trị x để P nhận giá trị nguyên.

b)Cho a, b, c là các số thực thõa mãn điều kiện abc\[ \ne \]0 và a+b+c\[ \ne \]0. Chứng minh rằng nếu \(\frac{{a + b}}{c} + \frac{{b + c}}{a} + \frac{{c + a}}{b} = - 2\)thì \(\frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}} + {b^{2023}} + {c^{2023}}}}.\)

a)Với \(x \ge 0,x \ne 4,x \ne 9\).

\(P = \left( {\frac{{3 - \sqrt x }}{{\sqrt x - 2}} + \frac{{\sqrt x - 2}}{{\sqrt x + 3}} - \frac{{9 - x}}{{x + \sqrt x - 6}}} \right)\) :\(\left( {\frac{{x - 3\sqrt x }}{{x - 9}} + \frac{{\sqrt x + 1}}{{x + 4\sqrt x + 3}}} \right)\)
\( = \left[ {\frac{{3 - \sqrt x }}{{\sqrt x - 2}} + \frac{{\sqrt x - 2}}{{\sqrt x + 3}} - \frac{{(3 - \sqrt x )(3 + \sqrt x )}}{{(\sqrt x - 2)(\sqrt x + 3)}}} \right]\):\(\left[ {\frac{{\sqrt x (\sqrt x - 3)}}{{(\sqrt x - 3)(\sqrt x + 3)}} + \frac{{\sqrt x + 1}}{{(\sqrt x + 1)(\sqrt x + 3)}}} \right]\)

\( = \frac{{\sqrt x - 2}}{{\sqrt x + 3}}:\frac{{\sqrt x + 1}}{{\sqrt x + 3}}\)

\( = \frac{{\sqrt x - 2}}{{\sqrt x + 1}} = 1 - \frac{3}{{\sqrt x + 1}}\)

Do \(x \ge 0 \Rightarrow \sqrt x + 1 \ge 1 \Rightarrow 1 - \frac{3}{{\sqrt x + 1}} \ge - 2\)

\(\frac{3}{{\sqrt x + 1}} > 0 \Rightarrow 1 - \frac{3}{{\sqrt x + 1}} < 1\)

Nên -2\[ \le \]P<1

\[P \in Z \Rightarrow \]{-2;-1;0}

\[P = - 2 \Rightarrow 1 - \frac{3}{{\sqrt x + 1}} = - 2 \Leftrightarrow \sqrt x + 1 = 1 \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0\](Thỏa mãn)

\[P = - 1 \Rightarrow 1 - \frac{3}{{\sqrt x + 1}} = - 1 \Leftrightarrow \sqrt x + 1 = \frac{3}{2} \Leftrightarrow \sqrt x = \frac{1}{2} \Leftrightarrow x = \frac{1}{4}\]( Thỏa mãn)

\[P = 0 \Rightarrow 1 - \frac{3}{{\sqrt x + 1}} = 0 \Leftrightarrow \sqrt x + 1 = 3 \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4\](Không thỏa mãn)

Vậy \[x \in \]{0; \[\frac{1}{4}\]}

b)Từ abc \[ \ne \]0 \[ \Rightarrow \]a,b,c \[ \ne \]0

\(\begin{array}{l}\frac{{a + b}}{c} + \frac{{b + c}}{a} + \frac{{c + a}}{b} = - 2 \Leftrightarrow \frac{{a + b}}{c} + 1 + \frac{{b + c}}{a} + 1 + \frac{{c + a}}{b} + 1 = 1\\ \Leftrightarrow \frac{{a + b + c}}{c} + \frac{{a + b + c}}{a} + \frac{{a + b + c}}{b} = 1\end{array}\)

\[\begin{array}{l} \Leftrightarrow (a + b + c)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) = 1\\ \Leftrightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{a + b + c}}(Do\;\;a + b + c \ne 0)\\ \Leftrightarrow \frac{{a + b}}{{ab}} = \frac{1}{{a + b + c}} - \frac{1}{c}\\ \Leftrightarrow \frac{{a + b}}{{ab}} = \frac{{ - (a + b)}}{{c(a + b + c)}}\end{array}\]

\[ \Leftrightarrow \]\(\left[ {\begin{array}{*{20}{c}}{a + b = 0}\\{ab = - c\left( {a + b + c} \right)}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{a = - b}\\{c + ab + ac + bc = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{a = - b}\\{\left( {c + a} \right)\left( {c + b} \right) = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{a = - b \ne 0}\\{c = - a \ne 0}\\{c = - b \ne 0}\end{array}} \right.\)

TH1:\(\;a = - b \ne 0\) \[ \Leftrightarrow \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}}}} + \frac{1}{{ - {a^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}} - {a^{2023}} + {c^{2023}}}} = \frac{1}{{{a^{2023}} + {b^{2023}} + {c^{2023}}}}\]

Tương tự cho TH2, TH3 \[ \Rightarrow \]điều phải chứng minh