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: \(-\dfrac{2}{3}.\)
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 \(\dfrac{1}{1} \ne \dfrac{5}{4}.\)
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 \(D = 0:\)
\(D = 64{a^2}-100{a^2}-48a-16 = 0\;\;\;\; \Leftrightarrow \;\;\;\;-36{a^2}-48a-16 = 0\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;9{a^2} + 12a + 4 = 0\;\;\;\; \Leftrightarrow \;\;\;\;{\left( {3a + 2} \right)^2} = 0\;\;\;\; \Leftrightarrow \;\;\;\;a = -\dfrac{2}{3}.\)
Ответ: \(-\dfrac{2}{3}.\)