a) Rút gọn biểu thức A= 8+ x ( 1+ căn bậc hai x -2 căn bậc hai x + 1/ (x -4 ) ( x - 2 căn bậc hai x + 4)

a) Rút gọn biểu thức \(A = \frac{{8 + x\left( {1 + \sqrt {x - 2\sqrt x  + 1} } \right)}}{{\left( {x - 4} \right)\left( {x - 2\sqrt x  + 4} \right)}} + \frac{{x - 3\sqrt x }}{{2\left( {x - \sqrt x  - 6} \right)}}\) (với \(x > 1,x \ne 4,x \ne 9\))

Với \(x > 1,x \ne 4,x \ne 9\)ta có:

\[\begin{array}{l}A = \frac{{8 + x\left( {1 + \sqrt {x - 2\sqrt x  + 1} } \right)}}{{\left( {x - 4} \right)\left( {x - 2\sqrt x  + 4} \right)}} + \frac{{x - 3\sqrt x }}{{2\left( {x - \sqrt x  - 6} \right)}}\\\,\,\,\,\, = \frac{{8 + x\left( {1 + \sqrt {{{\left( {\sqrt x  - 1} \right)}^2}} } \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)\left( {x - 2\sqrt x  + 4} \right)}} + \frac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \frac{{8 + x\left( {1 + \left| {\sqrt x  - 1} \right|} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {x\sqrt x  + 8} \right)}} + \frac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \frac{{8 + x\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {x\sqrt x  + 8} \right)}} + \frac{{\sqrt x }}{{2\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \frac{1}{{\left( {\sqrt x  - 2} \right)}} + \frac{{\sqrt x }}{{2\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \frac{{2\left( {\sqrt x  + 2} \right) + \sqrt x \left( {\sqrt x  - 2} \right)}}{{2\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \frac{{x + 4}}{{2\left( {x - 4} \right)}}\end{array}\]

b) Tìm tất cả bộ ba số nguyên tố \(p,q,r\)thỏa mãn \(pq = r + 1\) và \(2\left( {{p^2} + {q^2}} \right) = {r^2} + 1\).

Đặt \(\left\{ \begin{array}{l}S = p + q\\P = p.q\end{array} \right.\)ta có hệ:

\(\left\{ \begin{array}{l}P = r + 1\\2\left( {{S^2} - 2P} \right) = {r^2} + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}P = r + 1\\2\left( {{S^2} - 2P} \right) = {r^2} + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}P = r + 1\\{S^2} = \frac{{{r^2} + 4r + 5}}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}P = r + 1\\S = \sqrt {\frac{{{r^2} + 4r + 5}}{2}} \end{array} \right.\)

Vì \(p,q,r\) là ba số nguyên tố nên ta có:

\(\left\{ \begin{array}{l}r = 5\\S = 5\\P = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}r = 5\\p + q = 5\\p.q = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}r = 5\\q = 5 - p\\p.\left( {5 - p} \right) = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}r = 5\\q = 5 - p\\{p^2} - 5p + 6 = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}r = 5\\p = 2\\q = 3\end{array} \right.\)hoặc \(\left\{ \begin{array}{l}r = 5\\p = 3\\q = 2\end{array} \right.\).