Рассмотрим примеры замены функции, используя схему.
- $$ \log _{x^2 } \left( {5 + x} \right) \leftrightarrow \left( {x^2 - 1} \right)\left( {\left( {5 + x} \right) - 1} \right) = \left( {x^2 - 1} \right)\left( {4 + x} \right)$$, ОДЗ: $$
x > - 5,\quad x \ne \pm 1$$
- $$
\left| {x - 2} \right| - 4 - x^2 = \left| {x - 2} \right| - \left( {4 + x^2 } \right) \leftrightarrow \left( {x - 2} \right)^2 - \left( {4 + x^2 } \right)^2
$$, так как $$
\left| {x - 2} \right| \ge 0$$ Ъ $$
\left( {4 + x^2 } \right) \ge 0$$
- $$
\left| {x + 3} \right| - \sqrt {x^2 - x - 2} \leftrightarrow \left( {x + 3} \right)^2 - \left( {\sqrt {x^2 - x - 2} } \right)^2$$, так как $$
\left| {x + 3} \right| \ge 0{\rm{ ; }}\sqrt {x^2 - x - 2} \ge 0$$
- $$
8 - x^3 = 2^3 - x^3 \leftrightarrow 2 - x$$
- $$
\sqrt {x + 2} - \sqrt {2x + 3} \leftrightarrow \left( {x + 2} \right) - \left( {2x + 3} \right)$$, ОДЗ: $$
\left\{ \begin{array}{l}
x \ge - 2; \\
x \ge - 1,5. \\
\end{array} \right.$$
- $$
\left( {3^x - 1} \right) \leftrightarrow x \cdot \left( {3 - 1} \right) = 2x$$
- $$
\left| x \right|^{3x - 1} - \left| x \right|^{5 - x} \leftrightarrow \left( {\left( {3x - 1} \right) - \left( {5 - x} \right)} \right)\left( {\left| x \right| - 1} \right) \leftrightarrow \left( {4x - 6} \right)\left( {x^2 - 1} \right)$$, ОДЗ: $$
x \ne 0,\quad x \ne \pm 1.$$
- $$
\log _{x + 2} \left( {1 - \left| x \right|} \right) - \log _{x + 2} \left( {2 - 3\left| x \right|} \right)
\leftrightarrow \left( {\left( {1 - \left| x \right|} \right) - \left( {2 - 3\left| x \right|} \right)} \right)\left( {x + 2 - 1} \right)$$ = $$
\left( {2\left| x \right| - 1} \right)\left( {x + 1} \right) \leftrightarrow \left( {4x^2 - 1} \right)\left( {x + 1} \right)$$, ОДЗ: $$
\left\{ \begin{array}{l}
\left| x \right| < 1, \\
\left| x \right| < \frac{2}{3}, \\
x > - 2,\;x \ne - 1. \\
\end{array} \right.$$
- $$
\log _5^3 x^2 \leftrightarrow \log _5 x^2 = \log _5 x^2 - \log _5 1 = \left( {x^2 - 1} \right)\left( {5 - 1} \right)$$, ОДЗ: $$
x \ne 0$$
- $$
\left| {x + 1} \right| - \sqrt {5 - 2x - 2x^2 } = \sqrt {\left| {x + 1} \right|^2 } - \sqrt {5 - 2x - 2x^2 }$$$$
\leftrightarrow \left( {\left( {x + 1} \right)^2 - \left( {5 - 2x - 2x^2 } \right)} \right)$$, ОДЗ: $$
5 - 2x - 2x^2 \ge 0$$