Tính giá trị của biểu thức M = x⁵ – 6x³ + x tại \(x = \frac{{3 + \sqrt 2 }}{{2\sqrt 2 - 1}}\).

Hướng dẫn giải

Đáp án: 0

Ta có: \(x = \frac{{3 + \sqrt 2 }}{{2\sqrt 2 - 1}} = \frac{{\left( {3 + \sqrt 2 } \right)\left( {2\sqrt 2 + 1} \right)}}{{\left( {2\sqrt 2 - 1} \right)\left( {2\sqrt 2 + 1} \right)}} = \frac{{6\sqrt 2 + 3 + 4 + \sqrt 2 }}{7} = \frac{{7\sqrt 2 + 7}}{7} = \sqrt 2 + 1\).

Ta có: M = x⁵ – 6x³ + x

               \( = {\left( {\sqrt 2 + 1} \right)^5} - 6.{\left( {\sqrt 2 + 1} \right)^3} + \left( {\sqrt 2 + 1} \right)\)

               \( = \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^4} - 6{{\left( {\sqrt 2 + 1} \right)}^2} + 1} \right]\)

               \( = \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^4} - 2{{\left( {\sqrt 2 + 1} \right)}^2} + 1 - 4{{\left( {\sqrt 2 + 1} \right)}^2}} \right]\)

               \( = \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^4} - 2{{\left( {\sqrt 2 + 1} \right)}^2} + 1 - 4{{\left( {\sqrt 2 + 1} \right)}^2}} \right]\)

                \( = \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {{{\left( {\sqrt 2 + 1} \right)}^2} - 1} \right)}^2} - 4{{\left( {\sqrt 2 + 1} \right)}^2}} \right]\)

                \( = \left( {\sqrt 2 + 1} \right)\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} - 2\left( {\sqrt 2 + 1} \right) - 1} \right]\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} + 2\left( {\sqrt 2 + 1} \right) - 1} \right]\)

                \( = \left( {\sqrt 2 + 1} \right)\left[ {2 + 2\sqrt 2 + 1 - 2\sqrt 2 - 2 - 1} \right]\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} + 2\left( {\sqrt 2 + 1} \right) - 1} \right]\)

                \( = 0.\left( {\sqrt 2 + 1} \right).\left[ {{{\left( {\sqrt 2 + 1} \right)}^2} + 2\left( {\sqrt 2 + 1} \right) - 1} \right]\)

                = 0.