28В. Найдите все значения параметра a, при каждом из которых система уравнений  \(\left\{ {\begin{array}{*{20}{c}}{x + 5y = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + 8xy + 16{y^2}-8\,a\,x-32\,a\,y + 25{a^2} + 12a + 4 = 0}\end{array}} \right.\)  имеет единственное решение.

Ответ

ОТВЕТ:  \(-\frac{2}{3}.\)

Решение

\(\left\{ {\begin{array}{*{20}{c}}{x + 5y = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + 8xy + 16{y^2}-8\,a\,x-32\,a\,y + 25{a^2} + 12a + 4 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x + 5y = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {x + 4y} \right)}^2}-8\,a\,\left( {x + 4y} \right) + 25{a^2} + 12a + 4 = 0.}\end{array}} \right.\)

Пусть  \(x + 4y = t.\)  Тогда полученная система примет вид:

\(\left\{ {\begin{array}{*{20}{c}}{x + 5y = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{t^2}-8\,a\,t + 25{a^2} + 12a + 4 = 0.}\end{array}} \right.\)

При этом система  \(\left\{ {\begin{array}{*{20}{c}}{x + 5y = 3,}\\{x + 4y = t\;}\end{array}} \right.\)  будет иметь одно решение при \(t \in R,\) так как  \(\frac{1}{1} \ne \frac{5}{4}.\)

Следовательно, исходная система будет иметь одно решение, если уравнение  \({t^2}-8\,a\,t + 25{a^2} + 12a + 4 = 0\)  будет иметь один корень, а это возможно только в случае, если  \(D = 0:\)

\(D = 64{a^2}-100{a^2}-48a-16 = 0\;\;\;\; \Leftrightarrow \;\;\;\;-36{a^2}-48a-16 = 0\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;9{a^2} + 12a + 4 = 0\;\;\;\; \Leftrightarrow \;\;\;\;{\left( {3a + 2} \right)^2} = 0\;\;\;\; \Leftrightarrow \;\;\;\;a = -\frac{2}{3}.\)

Ответ:  \(-\frac{2}{3}.\)