Question

Va rog urgent dau coroana!!cat se poate de repede!!!​

Answer 1

Răspuns:

$a) |2 - \frac{2}{3} | = | \frac{6}{3} - \frac{2}{3} | = | \frac{4}{3} | \\ b) | - \frac{5}{6} + 1 | = | - \frac{5}{6} + \frac{6}{6} | = | \frac{1}{6} | \\ c) |2.25 - 7.25 + 0.65| = | - 5 + 0.65| = | - 4.35| = 4.35 \\ d) | \sqrt{3} - 2 | \\ \sqrt{1} < \sqrt{3} < \sqrt{4} \\ 1 < \sqrt{3} < 2 \\ cum \: \sqrt{3} < 2 \: \: \: \: | \sqrt{3} - 2 | = - ( \sqrt{3} - 2) = - \sqrt{3} + 2 \\ e) | \sqrt{6} - \sqrt{2} | \\ \sqrt{1} < \sqrt{2} < \sqrt{4} \\ 1 < \sqrt{2} < 4 \\ \sqrt{4} < \sqrt{6} < \sqrt{9} \\ 2 < \sqrt{6} < 3 \\ cum \: \: \sqrt{6} > 2 \: si \: \sqrt{2} < 2 \: \: rezulta \: ca \: | \sqrt{6} - \sqrt{2} | = \sqrt{6} - \sqrt{2} \\ f) | \sqrt{12} - \sqrt{18} | \\ \sqrt{9} < \sqrt{12} < \sqrt{16} \\ 3 < \sqrt{12} < 4 \\ \sqrt{16} < \sqrt{18} < \sqrt{25} \\ 4 < \sqrt{18} < 5 \\ cum \: \sqrt{12} < 4 \: \: si \: \sqrt{18} > 4 \: \: rezulta \: ca \: | \sqrt{12} - \sqrt{18} | = - ( \sqrt{12} - \sqrt{18} ) = - \sqrt{12} + \sqrt{18} \\ g) |3 \sqrt{2} + 5 - 9| = |3 \sqrt{2} - 4 | = | \sqrt{18} - 4 | \\ \sqrt{16} < \sqrt{18} < \sqrt{25} \\ 4 < \sqrt{18} < 5 \\ cum \: \: \sqrt{18} > 4 \: \: rezulta \: ca \: \: | \sqrt{18} - 4| = \sqrt{18} - 4 \\ h) | - \sqrt{6} - 6 + 8 | = | - \sqrt{6} + 2 | \\ cum \: \sqrt{4} = 2 < \sqrt{6} \: \: \: \: \: | - \sqrt{6} + 2 | = - \sqrt{6} + 2$