Cho phương trình

Vì \[\Delta  = {b^2} - 4ac = {2^2} - 4.\left( { - \sqrt 2 } \right).3 = 4 + 12\sqrt 2  > 0\]

Nên phương trình có hai nghiệm phân biệt \[{x_1},\,{x_2}\].

Theo định lí Viète, ta có: \[S = {x_1} + {x_2} = \frac{{ - b}}{a} = \frac{{ - 2}}{{ - \sqrt 2 }} = \frac{1}{{\sqrt 2 }};P = {x_1}.{x_2} = \frac{c}{a} = \frac{3}{{ - \sqrt 2 }} =  - \frac{3}{{\sqrt 2 }}\]

Do đó \(A = \frac{{{x_2} + 1}}{{1 - {x_1}}} + \frac{{{x_1} + 1}}{{1 - {x_2}}}\)\[ = \frac{{\left( {{x_2} + 1} \right)\left( {1 - {x_2}} \right)}}{{\left( {1 - {x_1}} \right)\left( {1 - {x_2}} \right)}} + \frac{{\left( {{x_1} + 1} \right)\left( {1 - {x_1}} \right)}}{{\left( {1 - {x_2}} \right)\left( {1 - {x_1}} \right)}} = \frac{{{x_2} - x_2^2 + 1 - {x_2} + {x_1} - x_1^2 + 1 - {x_1}}}{{1 - {x_2} - {x_1} + {x_1}{x_2}}}\]

\[ = \frac{{2 - \left( {x_1^2 + x_2^2} \right)}}{{1 - \left( {{x_1} + {x_2}} \right) + {x_1}{x_2}}} = \frac{{2 - \left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right]}}{{1 - \left( {{x_1} + {x_2}} \right) + {x_1}{x_2}}}\]\[ = \frac{{2 - \left( {{S^2} - 2P} \right)}}{{1 - S + P}} = \frac{{2 - \left[ {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} - 2.\left( { - \frac{3}{{\sqrt 2 }}} \right)} \right]}}{{1 - \frac{1}{{\sqrt 2 }} + \left( { - \frac{3}{{\sqrt 2 }}} \right)}} = \frac{3}{2}\]