3В.  При каких значениях параметров а и b уравнение  \(\left( {a + b} \right)\left( {a\,b\,x-1} \right) = a\,b + 6x-5\)  имеет более одного решения?

Ответ

ОТВЕТ:  \(a = 2,\,\,\,b = 1;\,\,\,\,a = 1,\,\,\,\,b = 2.\)

Решение

\(\left( {a + b} \right)\left( {a\,b\,x-1} \right) = a\,b + 6x-5\;\;\;\; \Leftrightarrow \;\;\;\;\left( {a + b} \right)a\,b\,x-6x = a\,b + a + b-5\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\left( {\left( {a + b} \right)a\,b\,-6} \right)x = a\,b + a + b-5.\)

Последнее уравнение будет иметь более одного решения, если:  \(\left\{ {\begin{array}{*{20}{c}}{\left( {a + b} \right)a\,b\,-6 = 0,}\\{a\,b + a + b-5 = 0.}\end{array}} \right.\)

Пусть  \(a + b = t,\;\;\;\;ab = z.\)  Тогда:

\(\left\{ {\begin{array}{*{20}{c}}{t\,z\,-6 = 0,\;\;}\\{z + t-5 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{{t^2}-5t\, + 6 = 0,}\\{z = 5-t\;\;\;\;\;\;\;\,\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t = 2,\;\;\;\,}\\{z = 5-t}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{t = 3,\;\;\;\;}\\{z = 5-t}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t = 2,}\\{z = 3\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{t = 3,\,}\\{z = 2.}\end{array}} \right.}\end{array}} \right.\)

Вернёмся к прежним переменным:

\(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{a + b = 2,}\\{ab = 3\;\;\;\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{a + b = 3,}\\{ab = 2\;\;\;\,}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{b = 2-a,\;\;\;\;}\\{a\left( {2-a} \right) = 3}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{b = 3-a,\;\;\;\;}\\{a\left( {3-a} \right) = 2}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{b = 2-a,\;\;\;\;\,\;\;\;}\\{{a^2}-2a + 3 = 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{b = 3-a,\;\;\;\;\;\,\;\;}\\{{a^2}-3a + 2 = 0}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\emptyset ,\;\;\;\;\;\;\;\;\;\;\;}\\{\left\{ {\begin{array}{*{20}{c}}{b = 3-a,}\\{\left[ {\begin{array}{*{20}{c}}{a = 2,}\\{a = 1\;\,}\end{array}\;\;\;} \right.}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{a = 2,}\\{b = 1\;\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{a = 1,\,}\\{b = 2.}\end{array}} \right.}\end{array}} \right.\)

Ответ:  \(a = 2,\,\,\,b = 1;\,\,\,\,a = 1,\,\,\,\,b = 2.\)