Cho x + y = 10; xy = 4. Tính a) x^2 + y^2 b) x^3 + y^3 c) x^4 + y^4 d) x^5 + y^5

Đáp án:

Giải thích các bước giải: ta có: $\begin{array}{l}a)\,\,{x^2} + {y^2} = {x^2} + {y^2} + 2xy - 2xy\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {x + y} \right)^2} - 2xy\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^2} - 2.4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 92\end{array}$ $\begin{array}{l}b)\,\,{x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\ = \left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} - 3xy} \right]\\ = 10.\left( {{{10}^2} - 3.4} \right)\\ = 10.88\\ = 880\end{array}$ $\begin{array}{l}c)\,{x^4} + {y^4} = {\left( {{x^2}} \right)^2} + 2{x^2}{y^2} + {\left( {{y^2}} \right)^2} - 2{x^2}{y^2}\\ = {\left( {{x^2} + {y^2}} \right)^2} - 2xy.xy\\ = {92^2} - 2.4.4\\ = 8432\end{array}$