Rovnica s absolútnou hodnotou
Rieš v \( \R \) rovnicu:
\( \normalsize\left|x-x^2-1\right|=\left|2x-3-x^2\right| \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = 1 \)
\( \normalsize K=\left\{1\right\} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = 2 \)
\( \normalsize K=\left\{2\right\} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = -1 \)
\( \normalsize K=\left\{-1\right\} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = 0 \)
\( \normalsize K=\left\{0\right\} \)
Výraz obsahuje dve absolútne hodnoty. Najprv teda určíš nulové body a rovnice vyriešiš pre vybrané intervaly.