a) Cho các số thực x,y khác 0, thoả mãn:

a) Từ giả thiết ta có

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{{x^2} + {y^2} = 3xy}\\{{x^3} + {y^3} = 10xy}\end{array}} \right. \Rightarrow \left( {x + y} \right)\left( {{x^2} + {y^2}} \right) = 3xy\left( {x + y} \right) \Rightarrow {x^3} + {y^3} + xy\left( {x + y} \right) = 3xy\left( {x + y} \right)\\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt}  \Leftrightarrow 10xy = 2xy\left( {x + y} \right) \Leftrightarrow x + y = 5\left( {x,y \ne 0} \right)\end{array}\)                                                 Ta có \({\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy = 5xy \Rightarrow xy = 5 \Rightarrow \frac{{x + y}}{{xy}} = \frac{1}{x} + \frac{1}{y} = 1\)(đpcm)

a)     Theo định lý Bezout: \(P\left( x \right) - 6 = S\left( x \right)\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\)

Do P bậc 3 \( \Rightarrow S\left( x \right) = a\). và \(P\left( { - 1} \right) = a\left( { - 2} \right)\left( { - 3} \right)\left( { - 4} \right) + 6 =  - 18 \Rightarrow a = 1\)

Suy ra \(P\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) + 6 = {x^3} - 6{x^2} + 11x\)

Thử lại ta thấy đúng.

Vậy \(P\left( x \right) = {x^3} - 6{x^2} + 11x\)

c) Đặt \(\left( {\sqrt a ,\sqrt b ,\sqrt c } \right) = \left( {x,y,z} \right)\)điều kiện: \(x,y,z \ge 0\)

    \( \Rightarrow x + y + z = 8;{x^2} + {y^2} + {z^2} = 26;{x^2}{y^2}{z^2} = 144\)      

\( \Rightarrow x + y + z = 8;xy + yz + zx = \frac{{{{\left( {x + y + z} \right)}^2} - \left( {{x^2} + {y^2} + {z^2}} \right)}}{2} = 19;xyz = 12\)(Do \(x,y,z \ge 0\))

Ta có:\(P = \frac{1}{{yz - x + 9}} + \frac{1}{{xz - y + 9}} + \frac{1}{{xy - z + 9}}\)                             

Ta có: \(yz - x + 9 = yz - x + x + y + z + 1 = \left( {z + 1} \right)\left( {y + 1} \right)\)

Tương tự: \(xz - y + 9 = \left( {x + 1} \right)\left( {z + 1} \right);xy - z + 9 = \left( {x + 1} \right)\left( {y + 1} \right)\)

\( \Rightarrow \frac{{x + 1 + y + 1 + z + 1}}{{\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right)}} = \)\(\frac{{x + y + z + 3}}{{xyz + x + y + z + xy + yz + xz + 1}}\) = \(\frac{{11}}{{12 + 19 + 8 + 1}}\) = \(\frac{{11}}{{40}}\)

Vậy P = \(\frac{{11}}{{40}}\)