ОГЭ по математике. №20 Уравнения. ФИПИ. Задача 3

ОТВЕТ:  \(-\dfrac{1}{2};\,\,\,\,\dfrac{1}{3}.\)

1 способ

Запишем область допустимых значений:  \(x \ne 0.\)

\(\dfrac{1}{{{x^2}}}-\dfrac{1}{x}-6 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{1-x-6{x^2}}}{{{x^2}}} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}1-x-6{x^2} = 0,\\x \ne 0.\end{array} \right.\)

\(1-x-6{x^2} = 0\left| { \cdot \left( {-1} \right)} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,6{x^2} + x-1 = 0;\,\,\,\,\,\,\,D = {1^2}-4 \cdot 6 \cdot \left( {-1} \right) = 25.\)

\(\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{-1 + 5}}{{2 \cdot 6}} = \dfrac{1}{3},\,\,\,\,}\\{x = \dfrac{{-1-5}}{{2 \cdot 6}} = -\dfrac{1}{2},}\end{array}} \right.\\x \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{1}{3},\\x = -\dfrac{1}{2}.\end{array} \right.\)

Ответ:  \(-\dfrac{1}{2};\,\,\,\,\dfrac{1}{3}.\)

2 способ

Запишем область допустимых значений:  \(x \ne 0.\)

Пусть  \(\dfrac{1}{x} = t,\)  тогда уравнение примет вид:

\({t^2}-t-6 = 0;\,\,\,\,\,\,D = {\left( {-1} \right)^2}-4 \cdot 1 \cdot \left( {-6} \right) = 25;\,\,\,\,\,\,\,\left[ \begin{array}{l}t = \dfrac{{1 + 5}}{2} = 3,\\t = \dfrac{{1-5}}{2} = -2.\end{array} \right.\)

Возвращаясь к прежней переменной, получим:

\(\left[ \begin{array}{l}\dfrac{1}{x} = 3,\\\dfrac{1}{x} = -2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{1}{3},\\x = -\dfrac{1}{2}.\end{array} \right.\)

Ответ:  \(-\dfrac{1}{2};\,\,\,\,\dfrac{1}{3}.\)