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Функције – нуле функције 2


Задаци


Текст задатака објашњених у видео лекцији:

Одредити нуле функције:

пр.4)  $y={\log _2}\frac{{{x^2}-9x + 18}}{{{x^2} - 5x+6}}$

 

 


пр.4)

\[\begin{gathered}
y = {\log _2}\frac{{{x^2} - 9x + 18}}{{{x^2} - 5x + 6}} \hfill \\
\begin{array}{*{20}{c}}
{Df:}&{\frac{{{x^2} - 9x + 18}}{{{x^2} - 5x + 6}} > 0}& \cap &{{x^2} - 5x + 6 \ne 0} \\
{}&{{x^2} - 9x + 18 = 0}&{}&{\begin{array}{*{20}{c}}
{{x_1} \ne 2}&{{x_2} \ne 3}
\end{array}} \\
{}&{{x_{1,2}} = \frac{{9 \pm \sqrt {81 - 72} }}{2}}&{}&{} \\
{}&{\begin{array}{*{20}{c}}
{{x_1} = 3}&{{x_2} = 6}
\end{array}}&{}&{} \\
{}&{{x^2} - 5x + 6 = 0}&{}&{} \\
{}&{{x_{1,2}} = \frac{{5 \pm \sqrt {25 - 24} }}{2}}&{}&{} \\
{}&{\begin{array}{*{20}{c}}
{{x_1} = 2}&{{x_2} = 3}
\end{array}}&{}&{}
\end{array} \hfill \\
\hfill \\
\end{gathered} \]425 png

426 png

\[Df:x \in \left( { - \infty ;2} \right) \cup \left( {6; + \infty } \right)\]

нуле:

\[\begin{gathered}
y = 0 \hfill \\
\begin{array}{*{20}{c}}
{{{\log }_2}\frac{{{x^2} - 9x + 18}}{{{x^2} - 5x + 6}} = 0}&{}&{\log A = 0 \Leftrightarrow A = 1}
\end{array} \hfill \\
\frac{{{x^2} - 9x + 18}}{{{x^2} - 5x + 6}} = 1 \hfill \\
\frac{{{x^2} - 9x + 18}}{{{x^2} - 5x + 6}} - 1 = 0 \hfill \\
\frac{{{x^2} - 9x + 18 - \left( {{x^2} - 5x + 6} \right)}}{{{x^2} - 5x + 6}} = 0 \hfill \\
\frac{{{x^2} - 9x + 18 - {x^2} + 5x - 6}}{{{x^2} - 5x + 6}} = 0 \hfill \\
\frac{{ - 4x + 12}}{{{x^2} - 5x + 6}} = 0 \hfill \\
- 4x + 12 = 0 \hfill \\
x = 3 \notin Df \hfill \\
\end{gathered} \]

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