Question

Ajutati-ma va rog!!!​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Explicație pas cu pas:

1)

$1 + 2 + 3 + ... + 24 = \frac{24\cdot 25}{2}$

$\frac{1}{12} + \frac{1}{2\cdot 3} + ... + \frac{1}{24\cdot 25} = \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{24} - \frac{1}{25} \\ = 1 - \frac{1}{25} = \frac{25 - 1}{25} = \frac{24}{25}$

→

$a = \sqrt{2\cdot \frac{24\cdot 25}{2}\cdot \frac{24}{25}} = \sqrt{ {24}^{2} } = 24$

2)

$\frac{1}{2}\cdot \frac{2}{3}\cdot  \frac{3}{4}\cdot...\cdot  \frac{81}{82} = \frac{1}{82}$

$\frac{1}{12} + \frac{1}{2\cdot 3} + ... + \frac{1}{81\cdot 82} = \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{81} - \frac{1}{82} \\ = 1 - \frac{1}{82} = \frac{82 - 1}{82} = \frac{81}{82}$

→

$b = \sqrt{ \frac{1}{82}\cdot  \frac{81}{82} } = \sqrt{ \frac{ {9}^{2} }{ {82}^{2} } } = \frac{9}{82}$

3)

$\frac{ \sqrt{2} - 1}{ \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{6} } + \frac{ \sqrt{4} - \sqrt{3} }{ \sqrt{12} } + ... + \frac{ \sqrt{64} - \sqrt{63}}{ \sqrt{4032} } = \\ = \frac{ \sqrt{2}}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } + \frac{ \sqrt{3}}{ \sqrt{6} } - \frac{ \sqrt{2}}{ \sqrt{6} } + \frac{ \sqrt{4}}{ \sqrt{12} } - \frac{\sqrt{3} }{ \sqrt{12} } + ... + \frac{ \sqrt{64}}{ \sqrt{63\cdot64} } - \frac{\sqrt{63}}{ \sqrt{63\cdot64} } \\ = 1 - \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{3} } + \frac{1}{ \sqrt{3} } - \frac{1}{ \sqrt{4} } + ... + \frac{1}{ \sqrt{63} } - \frac {1}{ \sqrt{64} } \\ = 1 - \frac {1}{ \sqrt{64} } = 1 - \frac{1}{8} = \frac{8 - 1}{8} = \frac{7}{8}$