Cho biểu thức \(P = \frac{{3x + 5\sqrt x  - 11}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}} - \frac{{\sqrt x  - 2}}{{\sqrt x  - 1}} + \frac{2}{{\sqrt x  + 2}} - 1\)   (với \(0 \le x \ne 1\))

a) Rút gọn biểu thức \[P\].

b) Tìm \[x\] để \[P\] chia hết cho 3.

Với \(n \in \mathbb{N}\), ta có \({n^5} + 1 \vdots {n^3} + 1 \Leftrightarrow {n^2}\left( {{n^3} + 1} \right) - \left( {{n^2} - 1} \right) \vdots {n^3} + 1\)

\( \Leftrightarrow \left( {{n^2} - 1} \right) \vdots {n^3} + 1 \Leftrightarrow \left( {n - 1} \right)\left( {n + 1} \right) \vdots \left( {n + 1} \right)\left( {{n^2} - n + 1} \right)\)

\( \Leftrightarrow n - 1 \vdots {n^2} - n + 1\) (vì \(n + 1 \ne 0\))

\( \Rightarrow n\left( {n - 1} \right) \vdots {n^2} - n + 1 \Leftrightarrow \left( {{n^2} - n + 1} \right) - 1 \vdots {n^2} - n + 1\)

\( \Leftrightarrow 1 \vdots {n^2} - n + 1 \Rightarrow \left[ \begin{array}{l}{n^2} - n + 1 = 1\\{n^2} - n + 1 =  - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}n = 1\\n = 0\end{array} \right.\)

Thử lại ta thấy \(n = 0;\,{\rm{ }}n = 1\) thỏa mãn để \({n^5} + 1\) chia hết cho \({n^3} + 1\)

Vậy \(n = 0;\,{\rm{ }}n = 1.\)

Đặt \(\left\{ {\begin{array}{*{20}{c}}{x = a + b - c > 0}\\{y = b + c - a > 0}\\{z = c + a - b > 0}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{a = \frac{{x + z}}{2}}\\{b = \frac{{x + y}}{2}}\\{c = \frac{{y + z}}{2}}\end{array}} \right.\)

Ta cần chứng minh: \(\frac{{{{\left( {x + y} \right)}^2}}}{{4z}} + \frac{{{{\left( {y + z} \right)}^2}}}{{4x}} + \frac{{{{\left( {z + x} \right)}^2}}}{{4y}} \ge x + y + z\)

Ta có: \(\frac{{{{\left( {x + y} \right)}^2}}}{{4z}} + \frac{{{{\left( {y + z} \right)}^2}}}{{4x}} + \frac{{{{\left( {z + x} \right)}^2}}}{{4y}} \ge \frac{{xy}}{z} + \frac{{yz}}{x} + \frac{{zx}}{y}{\rm{  }}\left( 1 \right)\)

Mặt khác: \(\frac{{xy}}{z} + \frac{{yz}}{x} \ge 2y;{\rm{ }}\frac{{yz}}{x} + \frac{{zx}}{y} \ge 2z;{\rm{ }}\frac{{xy}}{z} + \frac{{zx}}{y} \ge 2x\).

Khi đó \(\frac{{xy}}{z} + \frac{{yz}}{x} + \frac{{zx}}{y} \ge x + y + z{\rm{  }}\left( 2 \right)\)

Từ \(\left( 1 \right){\rm{, }}\left( 2 \right)\) ta có \(\frac{{{{\left( {x + y} \right)}^2}}}{{4z}} + \frac{{{{\left( {y + z} \right)}^2}}}{{4x}} + \frac{{{{\left( {z + x} \right)}^2}}}{{4y}} \ge x + y + z\)

Vậy \(\frac{{{a^2}}}{{b + c - a}} + \frac{{{b^2}}}{{c + a - b}} + \frac{{{c^2}}}{{a + b - c}} \ge a + b + c\)

Dấu bằng xãy ra khi \[a = b = c\]