1. Tìm tất cả các số nguyên tố \(p\) lẻ sao cho \(2{p^4} - {p^2} + 16\)là số chính phương.

2. Tìm nghiệm nguyên của phương trình \(6{x^2} + 7xy + 2{y^2} + x + y - 2 = 0\).

1)Ta có: \({a^2} + {b^2} + 2 = {\left( {a + b} \right)^2}\)

Nên

\(\begin{array}{l}Q = \frac{1}{{{{\left( {a + b} \right)}^3}}}\left( {\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}}} \right) + \frac{3}{{{{\left( {a + b} \right)}^4}}}\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \right) + \frac{6}{{{{\left( {a + b} \right)}^4}}}\\ = \frac{{{a^3} + {b^3}}}{{{{\left( {a + b} \right)}^3}}} + \frac{{3\left( {{a^2} + {b^2}} \right)}}{{{{\left( {a + b} \right)}^4}}} + \frac{6}{{{{\left( {a + b} \right)}^4}}}\\ = \frac{{\left( {{a^3} + {b^3}} \right)\left( {a + b} \right) + 3\left( {{a^2} + {b^2}} \right) + 6}}{{{{\left( {a + b} \right)}^4}}}\end{array}\)

\(\begin{array}{l} = \frac{{{a^4} + {b^4} + ab\left( {{a^2} + {b^2}} \right) + 3\left( {{a^2} + {b^2}} \right) + 6}}{{{{\left( {{a^2} + {b^2} + 2} \right)}^2}}}\\ = \frac{{{a^4} + {b^4} + 4\left( {{a^2} + {b^2}} \right) + 6}}{{{{\left( {{a^2} + {b^2} + 2} \right)}^2}}}\end{array}\)

\[\begin{array}{l} = \frac{{\left( {{a^4} + {b^4} + 2{a^2}{b^2}} \right) + 4\left( {{a^2} + {b^2}} \right) + 4}}{{{{\left( {{a^2} + {b^2} + 2} \right)}^2}}}\\ = \frac{{{{\left( {{a^2} + {b^2}} \right)}^2} + 4\left( {{a^2} + {b^2}} \right) + 4}}{{{{\left( {{a^2} + {b^2} + 2} \right)}^2}}}\end{array}\]

\[\begin{array}{l} = \frac{{{{\left( {{a^2} + {b^2} + 2} \right)}^2}}}{{{{\left( {{a^2} + {b^2} + 2} \right)}^2}}}\\ = 1\end{array}\]

2)\(P = \sqrt {{x^2} + 1} \sqrt {{y^2} + 1} + xy - \left( {x\sqrt {{y^2} + 1} + y\sqrt {{x^2} + 1} } \right) = \sqrt {{x^2} + 1} \sqrt {{y^2} + 1} + xy - \sqrt {15} \)

Đặt \(\begin{array}{l}M = \sqrt {{x^2} + 1} \sqrt {{y^2} + 1} + xy \Rightarrow {M^2} = \left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right) + {x^2}{y^2} + 2xy\sqrt {{x^2} + 1} .\sqrt {{y^2} + 1} \\ = 2{x^2}{y^2} + {x^2} + {y^2} + 1 + 2xy\sqrt {{x^2} + 1} .\sqrt {{y^2} + 1} \end{array}\)

\(\begin{array}{l} = {x^2}\left( {{y^2} + 1} \right) + {y^2}\left( {{x^2} + 1} \right) + 2x\sqrt {{y^2} + 1} .y\sqrt {{x^2} + 1} + 1\\ = {\left( {x\sqrt {{y^2} + 1} + y\sqrt {{x^2} + 1} } \right)^2} + 1\end{array}\)

\( = 16 \Rightarrow M = 4\). Vậy \(P = 4 - \sqrt {15} \).

1)Điều kiện: \(\left\{ \begin{array}{l}{x^2} + 3x \ge 0\\x - 1 \ge 0\\\frac{{{x^2} + 2x - 3}}{x} \ge 0\end{array} \right. \Leftrightarrow x \ge 1\)

Phương trình trở thành

\(\begin{array}{l}\sqrt {x\left( {x + 3} \right)} + 2\sqrt {x - 1} - 2x - \sqrt {\frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{x}} = 0\\ \Leftrightarrow \left( {\sqrt {x\left( {x + 3} \right)} - \sqrt {\frac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{x}} } \right) + \left( {2\sqrt {x - 1} - 2x} \right) = 0\\ \Leftrightarrow \sqrt {\frac{{x + 3}}{x}} \left( {x - \sqrt {x - 1} } \right) - 2\left( {x - \sqrt {x - 1} } \right) = 0\\ \Leftrightarrow \left( {x - \sqrt {x - 1} } \right)\left( {\sqrt {\frac{{x + 3}}{x}}  - 2} \right) = 0\\\left[ \begin{array}{l}x - \sqrt {x - 1} = 0\\\sqrt {\frac{{x + 3}}{x}} - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \sqrt {x - 1} \,\,\left( 1 \right)\\\sqrt {\frac{{x + 3}}{x}} = 2\,\left( 2 \right)\end{array} \right.\end{array}\)

\(\left( 1 \right) \Leftrightarrow {x^2} = x - 1 \Leftrightarrow {x^2} - x + 1 = 0\) (vô nghiệm)

\(\left( 2 \right) \Leftrightarrow \frac{{x + 3}}{x} = 4 \Leftrightarrow x + 3 = 4x \Leftrightarrow x = 1\) (Thoả mãn điều kiện)

2)Hệ phương trình đã cho trở thành \(\left\{ \begin{array}{l}\left( {x + 1} \right)\left( {y + 2} \right) = 4\\{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} = 8\end{array} \right.\)

Đặt \(\left\{ \begin{array}{l}a = x + 1\\b = y + 2\end{array} \right.\) ta được hệ \(\left\{ \begin{array}{l}a.b = 4\\{a^2} + {b^2} = 8\end{array} \right.\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}ab = 4\\{\left( {a + b} \right)^2} - 2ab = 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}ab = 4\\{\left( {a + b} \right)^2} = 16\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = 4\\\left[ \begin{array}{l}a + b = 4\\a + b = - 4\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}ab = 4\\a + b = 4\end{array} \right.\,\,\,\left( 1 \right)\\\left\{ \begin{array}{l}ab = 4\\a + b = - 4\end{array} \right.\,\left( 2 \right)\end{array} \right.\end{array}\)

\(\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 2\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = 1\\y = 0\end{array} \right.\)

\(\left( 2 \right) \Leftrightarrow \left\{ \begin{array}{l}a = - 2\\b = - 2\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = - 3\\y = - 4\end{array} \right.\)