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Limbo206
@Limbo206
August 2022
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Пусть x1, x2 — корни уравнения 2x^2−4x+1=0. Найдите значение выражения x1*x2^3+x1^3*x2
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Удачник66
Verified answer
По теореме Виета
x1 + x2 = -b/a = 4/2 = 2
x1*x2 = c/a = 1/2
x1*x2^3 + x1^3*x2 = x1*x2*(x1^2 + x2^2) = x1*x2*(x1^2+2x1*x2+x2^2-2x1*x2) =
= x1*x2*((x1 + x2)^2 - 2x1*x2) = 1/2*(2^2 - 2*1/2) = 1/2*(4 - 1) = 3/2
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Answers & Comments
Verified answer
По теореме Виетаx1 + x2 = -b/a = 4/2 = 2
x1*x2 = c/a = 1/2
x1*x2^3 + x1^3*x2 = x1*x2*(x1^2 + x2^2) = x1*x2*(x1^2+2x1*x2+x2^2-2x1*x2) =
= x1*x2*((x1 + x2)^2 - 2x1*x2) = 1/2*(2^2 - 2*1/2) = 1/2*(4 - 1) = 3/2