ОГЭ по математике. №20 Уравнения. ФИПИ. Задача 2

ОТВЕТ:  \(-1;\,\,\,\,\dfrac{1}{4}.\)

1 способ

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

\(\dfrac{1}{{{x^2}}}-\dfrac{3}{x}-4 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{1-3x-4{x^2}}}{{{x^2}}} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}1-3x-4{x^2} = 0,\\x \ne 0.\end{array} \right.\)

\(1-3x-4{x^2} = 0\left| { \cdot \left( {-1} \right)} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,4{x^2} + 3x-1 = 0;\,\,\,\,\,\,\,D = {3^2}-4 \cdot 4 \cdot \left( {-1} \right) = 25.\)

\(\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{-3 + 5}}{{2 \cdot 4}} = \dfrac{1}{4},\,\,\,\,}\\{x = \dfrac{{-3-5}}{{2 \cdot 4}} = -1,}\end{array}} \right.\\x \ne 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{1}{4},\\x = -1.\end{array} \right.\)

Ответ:  \(-1;\,\,\,\,\dfrac{1}{4}.\)

2 способ

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

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

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

Возвращаясь к прежней переменной, получим:

\(\left[ \begin{array}{l}\dfrac{1}{x} = 4,\\\dfrac{1}{x} = -1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{1}{4},\\x = -1.\end{array} \right.\)

Ответ:  \(-1;\,\,\,\,\dfrac{1}{4}.\)