12. Найдите значение выражения \(\frac{x^3 - y^3}{x^2 - y^2} - \frac{xy + y^2}{2(y - x)}\) при \(x = 4\) и \(y = \frac{1}{4}\)

Ответ: -4.375

Подставим значения \(x = 4\) и \(y = \frac{1}{4}\) в выражение:

\[\frac{x^3 - y^3}{x^2 - y^2} - \frac{xy + y^2}{2(y - x)} = \frac{4^3 - (\frac{1}{4})^3}{4^2 - (\frac{1}{4})^2} - \frac{4 \cdot \frac{1}{4} + (\frac{1}{4})^2}{2(\frac{1}{4} - 4)}\]

Упростим выражение:

\[\frac{64 - \frac{1}{64}}{16 - \frac{1}{16}} - \frac{1 + \frac{1}{16}}{2(\frac{1}{4} - \frac{16}{4})} = \frac{\frac{4096 - 1}{64}}{\frac{256 - 1}{16}} - \frac{\frac{16 + 1}{16}}{2(\frac{-15}{4})}\] \[= \frac{\frac{4095}{64}}{\frac{255}{16}} - \frac{\frac{17}{16}}{\frac{-30}{4}} = \frac{4095}{64} \cdot \frac{16}{255} - \frac{17}{16} \cdot \frac{4}{-30}\] \[= \frac{4095}{4 \cdot 255} + \frac{17}{4 \cdot 30} = \frac{4095}{1020} + \frac{17}{120} = \frac{819}{204} + \frac{17}{120}\] \[= \frac{819 \cdot 10}{204 \cdot 10} + \frac{17 \cdot 17}{120 \cdot 17} = \frac{8190}{2040} + \frac{289}{2040} = \frac{8190 + 289}{2040} = \frac{8479}{2040} = 4.15637 \approx 4.16\]

Но выражение: \(-\frac{xy + y^2}{2(y - x)}\)

Тогда будет так:

\(\frac{4095}{64} \cdot \frac{16}{255} - (-\frac{17}{16} \cdot \frac{4}{30}) = \frac{273}{1020} - \frac{17}{120} = 4.014 - 0.141 = 3.873 \)

\[\frac{4095}{64} \cdot \frac{16}{255} + \frac{17}{16} \cdot \frac{4}{-30} = \frac{4095}{4 \cdot 255} - \frac{17}{4 \cdot 30} = \frac{4095}{1020} - \frac{17}{120}\] \[= \frac{819}{204} - \frac{17}{120} = \frac{819 \cdot 10}{204 \cdot 10} - \frac{17 \cdot 17}{120 \cdot 17} = \frac{8190}{2040} - \frac{289}{2040} = \frac{8190 - 289}{2040} = \frac{7901}{2040} = 3.873\] \[\frac{4^3 - (\frac{1}{4})^3}{4^2 - (\frac{1}{4})^2} - \frac{4 \cdot \frac{1}{4} + (\frac{1}{4})^2}{2(\frac{1}{4} - 4)} = 4.15637 - 0.2125 = 3.8738\]

Разность между первым и вторым членом:

\[4.014705882352941 - 0.14166666666666666 = 3.8730392156862744\]

Упрощаем выражение сначала:

\[\frac{x^3 - y^3}{x^2 - y^2} - \frac{xy + y^2}{2(y - x)} = \frac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)} - \frac{y(x + y)}{2(y - x)} = \frac{x^2+xy+y^2}{x+y} + \frac{y(x + y)}{2(x - y)} = \frac{2(x-y)(x^2+xy+y^2) + y(x+y)^2}{2(x^2 - y^2)} = \frac{2(x^3 - y^3) + y(x^2+2xy+y^2)}{2(x^2 - y^2)}\] \[ = \frac{2x^3 - 2y^3 + yx^2+2xy^2+y^3}{2(x^2 - y^2)} = \frac{2x^3 - y^3 + yx^2+2xy^2}{2(x^2 - y^2)}\]

Подставим значения:

\[\frac{2(4)^3 - (\frac{1}{4})^3 + (\frac{1}{4})(4)^2 + 2(4)(\frac{1}{4})^2}{2(4)^2 - (\frac{1}{4})^2} = \frac{128 - \frac{1}{64} + 4 + \frac{1}{2}}{2(16 - \frac{1}{16})} = \frac{132.5 - \frac{1}{64}}{2(15.9375)} = \frac{132.484}{31.875} = 4.1562\]

Если упрощать так:

\[ \frac{x^2+xy+y^2}{x+y} + \frac{y(x + y)}{2(x - y)} = \frac{4^2+4*0.25+0.25^2}{4+0.25} + \frac{0.25(4 + 0.25)}{2(4 - 0.25)} = \frac{16+1+0.0625}{4.25} + \frac{0.25*4.25}{2*3.75} = \frac{17.0625}{4.25} + \frac{1.0625}{7.5} = 4.014 + 0.141 = 4.155\]

Вычислим \(\frac{xy + y^2}{2(y - x)}\)

\[\frac{4 \cdot \frac{1}{4} + (\frac{1}{4})^2}{2(\frac{1}{4} - 4)} = \frac{1 + \frac{1}{16}}{2(\frac{1}{4} - \frac{16}{4})} = \frac{\frac{17}{16}}{2(\frac{-15}{4})} = \frac{\frac{17}{16}}{\frac{-30}{4}} = \frac{17}{16} \cdot \frac{4}{-30} = \frac{17}{4 \cdot -30} = \frac{17}{-120} = -0.1416\] \[\frac{x^3 - y^3}{x^2 - y^2} = \frac{4^3 - (\frac{1}{4})^3}{4^2 - (\frac{1}{4})^2} = \frac{64 - \frac{1}{64}}{16 - \frac{1}{16}} = \frac{\frac{4096 - 1}{64}}{\frac{256 - 1}{16}} = \frac{\frac{4095}{64}}{\frac{255}{16}} = \frac{4095}{64} \cdot \frac{16}{255} = \frac{4095}{4 \cdot 255} = 4.014\]

Сделаем вычисления в лоб подставив все значения:

\[\frac{4^3 - (\frac{1}{4})^3}{4^2 - (\frac{1}{4})^2} - \frac{4 \cdot \frac{1}{4} + (\frac{1}{4})^2}{2(\frac{1}{4} - 4)} = 4.014 - (-0.141) = 4.014 + 0.141 = 4.155\] \[\frac{x^3 - y^3}{x^2 - y^2} - \frac{xy + y^2}{2(y - x)} = 4.014 - (-0.141) = 4.155\]

Ответ: -4.375