Question

Ajutor!Multumesc anticipat!​

Answer 1

Explicație pas cu pas:

$[\sqrt{3} + \sqrt{25} ] = [\sqrt{3} + 5] = 5 + [ \sqrt{3} ]$

$\sqrt{1} < \sqrt{3} < \sqrt{4} \implies 1 < \sqrt{3} < 2$

atunci:

$[\sqrt{3} + \sqrt{25} ] = 5 + 1 = 6$

.

$[\sqrt{4} + \sqrt{19} ] = [2 + \sqrt{19}] = 2 + [ \sqrt{19} ]$

$\sqrt{16} < \sqrt{19} < \sqrt{25} \implies 4 < \sqrt{19} < 5$

atunci:

$[\sqrt{4} + \sqrt{19} ] = 2 + 4 = 6$

=>

$[\sqrt{3} + \sqrt{25} ] = [\sqrt{4} + \sqrt{19} ]$

q.e.d.