Řešení rovnice s absolutními hodnotami
Řeš v \( \R \) rovnici:
\( \large \left | x-x^{2}-1 \right | = \left | 2 x-3-x^{2} \right | \)
\( I. x \in \mathbb{R} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = -1 \)
\( I. x \in \mathbb{R} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = 1 \)
\( I. x \in \mathbb{R} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = 2 \)
\( I. x \in \mathbb{R} \)
\( \left| {x\ –\ x^{2}\ –\ 1} \right| = \left| {2 x\ –\ 3\ –\ x^{2}} \right| \)
\( – \left( {x\ –\ x^{2}\ –\ 1} \right) =\ – \left( {2 x\ –\ 3\ –\ x^{2}} \right) \)
\( x\ –\ x^{2}\ –\ 1 = 2 x\ –\ 3\ –\ x^{2}\ \ \ \ \ /:\left( { – 1} \right) \)
\( x^{2}\ –\ x + 1 = x^{2}\ –\ 2 x + 3 \)
\( x = 0 \)
Výraz obsahuje dvě absolutní hodnoty, tudíž určíš nulové body a rovnice vyřešíš pro vybrané intervaly.