Не вычисляя корней уравнения 3x2+8x−1=0, найдите x1x42+x2x41.

Ответ:

Пошаговое объяснение:

3x²+8x−1=0,

$x_{1} +x_{2}=-\frac{8}{3} \\\\x_{1} x_{2}=-\frac{1}{3} \\\\(x_{1} +x_{2})^2=x^2_{1} +2x_{1} x_{2} + x^2_{2} =\frac{64}{9} ;\\ \\x^2_{1} + x^2_{2}=\frac{64}{9} -2(-\frac{1}{3}) =\frac{64}{9} +\frac{2}{3}=\frac{64+6}{9} =\frac{70}{9} \\\\x_{1} x^4_{2}+x_{2} x^4_{1}=x_{1} x_{2}(x^3_{1} +x^3_{2})= x_{1} x_{2}(x_{1} +x_{2})(x^2_{1} -x_{1} x_{2}+x^2_{2})=\\\\=-\frac{1}{3}*(-\frac{8}{3})* (\frac{70}{9} +\frac{1}{3}) =\frac{8}{9} *\frac{73}{9}=\frac{584}{81} =7\frac{17}{81} \\\\$