Вычислите значение выражения х в кубе - у в кубе - 2у в квадрате - 3у - 1+ х в квадрате - 2ху,если х = 71,8 и у=70,8
x^3 - y^3 - 2y^2 - 3y - 1 + x^2 - 2xy =
(x^2 - 2xy + y^2) - 3y^2 - 3y - 1 + x^3 - y^3 =
(x-y)^2 - (y^3 + 3y^2 + 3y + 1) + x^3 =
(x-y)^2 - (y+1)^3 + x^3 =
(71.8 - 70.8)^2 - (70.8+1)^3 + (71.8)^3 = 1
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x^3 - y^3 - 2y^2 - 3y - 1 + x^2 - 2xy =
(x^2 - 2xy + y^2) - 3y^2 - 3y - 1 + x^3 - y^3 =
(x-y)^2 - (y^3 + 3y^2 + 3y + 1) + x^3 =
(x-y)^2 - (y+1)^3 + x^3 =
(71.8 - 70.8)^2 - (70.8+1)^3 + (71.8)^3 = 1