Question

11 si 14 pls maxim stele si puncte de la 11a si 14 b

Answer 1

$11.a) \: \: | - 3| + |3 - \sqrt{11} | + |4 - \sqrt{11} | \\ = 3 - (3 - \sqrt{11} ) + 4 - \sqrt{11} \\ = 3 - 3 + \sqrt{11} + 4 - \sqrt{11} \\ = 3 - 3 + 4 \\ = 0 + 4 \\ = 4 \\ \\ \\ 14.b) \: \: \sqrt{(3 - \sqrt{7}) {}^{2} } + \sqrt{( \sqrt{7} - 2) {}^{2} } \\ = 3 - \sqrt{7} + \sqrt{7} - 2 \\ = 3 - 2 \\ = 1$

Answer 2

Explicație pas cu pas:

11.a)

$3 - \sqrt{11} = \sqrt{9} - \sqrt{11} < 0 \\ \implies |3 - \sqrt{11}| = \sqrt{11} - 3 \\ 4 - \sqrt{11} = \sqrt{16} - \sqrt{11} > 0 \\ \implies |4 - \sqrt{11}| = 4 - \sqrt{11}$

=>

$| - 3| + |3 - \sqrt{11} | + |4 - \sqrt{11} | = 3 + \sqrt{11} - 3 + 4 - \sqrt{11} = \bf 4$

14.b)

$3 - \sqrt{7} = \sqrt{9} - \sqrt{7} > 0 \\ \implies |3 - \sqrt{7}| = 3 - \sqrt{7} \\ \sqrt{7} - 2 = \sqrt{7} - \sqrt{4} > 0 \\ \implies |\sqrt{7} - 2| = \sqrt{7} - 2$

=>

$\sqrt{ {(3 - \sqrt{7} )}^{2} } + \sqrt{ {( \sqrt{7} - 2)}^{2} } = |3 - \sqrt{7} | + | \sqrt{7} - 2| = 3 - \sqrt{7} + \sqrt{7} - 2 = \bf 1$