27В. Найдите все значения параметра a, при каждом из которых система уравнений \(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + 6xy + 9{y^2}-10\,a\,x-30\,a\,y + 125{a^2} + 60a + 9 = 0}\end{array}} \right.\) имеет единственное решение.
Ответ
ОТВЕТ: \(-\dfrac{3}{{10}}.\)
Решение
\(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + 6xy + 9{y^2}-10\,a\,x-30\,a\,y + 125{a^2} + 60a + 9 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {x + 3y} \right)}^2}-10\,a\,\left( {x + 3y} \right) + 125{a^2} + 60a + 9 = 0.}\end{array}} \right.\)
Пусть \(x + 3y = t.\) Тогда полученная система примет вид:
\(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{t^2}-10\,a\,t + 125{a^2} + 60a + 9 = 0.}\end{array}} \right.\)
При этом система \(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,}\\{x + 3y = t\;\;\;}\end{array}} \right.\) будет иметь одно решение при \(t \in R,\) так как \(\dfrac{1}{1} \ne \dfrac{{-3}}{3}.\)
Следовательно, исходная система будет иметь одно решение, если уравнение \({t^2}-10\,a\,t + 125{a^2} + 60a + 9 = 0\) будет иметь один корень, а это возможно только в случае, если \(D = 0:\)
\(D = 100{a^2}-500{a^2}-240a-36 = 0\;\;\;\; \Leftrightarrow \;\;\;\;-400{a^2}-240a-36 = 0\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;100{a^2} + 60a + 9 = 0\;\;\;\; \Leftrightarrow \;\;\;\;{\left( {10a + 3} \right)^2} = 0\;\;\;\; \Leftrightarrow \;\;\;\;a = -\dfrac{3}{{10}}.\)
Ответ: \(-\dfrac{3}{{10}}.\)