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Lösung Element von

Version 3.1 von akukin am 2024/04/01 16:43
\(\mathbb{N}\)\(\mathbb{N}_0\)\(\mathbb{Z}^-\)\(\mathbb{Z}_+\)\(\mathbb{Z}\)\(\mathbb{Q}^-\)\(\mathbb{Q}^+\)\(\mathbb{Q}\)\(\mathbb{R}^-\)\(\mathbb{R}^+\)\(\mathbb{R}\)
 \(\frac{3}{4}\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\in\)\(\in\)\(\notin\)\(\in\)\(\in\)
 \(\frac{-4}{5}\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\in\)\(\notin\)\(\in\)\(\in\)\(\notin\)\(\in\)
 \(-\frac{6}{5}\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\notin\)\(\in\)\(\notin\)\(\in\)\(\in\)\(\notin\)\(\in\)
 \(\frac{10}{2}=5|={{formula}}\in{{/formula}}|{{formula}}\in{{/formula}}|{{formula}}\notin{{/formula}}|{{formula}}\in{{/formula}}|={{formula}}\in{{/formula}}|{{formula}}\notin{{/formula}}|={{formula}}\in{{/formula}}|{{formula}}\in{{/formula}}|={{formula}}\notin{{/formula}}|{{formula}}\in{{/formula}}|={{formula}}\in{{/formula}} |= {{formula}}4{{/formula}}|={{formula}}\in{{/formula}}|{{formula}}\in{{/formula}}|{{formula}}\notin{{/formula}}|{{formula}}\in{{/formula}}|={{formula}}\in{{/formula}}|{{formula}}\notin{{/formula}}|={{formula}}\in{{/formula}}|{{formula}}\in{{/formula}}|={{formula}}\notin{{/formula}}|{{formula}}\in{{/formula}}|={{formula}}\in{{/formula}} |= {{formula}}0{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}-6{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}\sqrt[4]{16}{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}\sqrt{4}{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}\sqrt{5}{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}(-3)^5{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}3^{-1}{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}(-2)^{-2}{{/formula}}|=|=|=|=|=|=|=|=|=|=|= |= {{formula}}tan 45^{o}{{/formula}}|=|=|=|=|=|=|=|=|=|=|=\)