Question

vaaa roggg!!!!! DAU COROANA!!!!​

Answer 1

Explicație pas cu pas:

b)

$\sqrt{ \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot...\cdot \frac{81}{82} \left( \frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{80 \cdot 81} + \frac{1}{81 \cdot 82}\right)} \\ = \sqrt{ \frac{1}{82} \left( \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{80} - \frac{1}{81} + \frac{1}{81} - \frac{1}{82} \right)} \\ = \sqrt{ \frac{1}{82} \left( \frac{1}{1} - \frac{1}{82} \right)} = \sqrt{ \frac{1}{82} \left(\frac{81}{82} \right)} = \sqrt{ \frac{ {9}^{2} }{ {82}^{2} }} = \frac{9}{82}$

c)

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

$= \left( \frac{ \sqrt{2} }{ \sqrt{2} } - \frac{1}{ \sqrt{2} } \right) +  \left( \frac{ \sqrt{3} }{ \sqrt{2 \times 3} } - \frac{ \sqrt{2} }{ \sqrt{2 \times 3} } \right) + \left( \frac{ \sqrt{4} }{ \sqrt{3 \times 4} } - \frac{ \sqrt{3} }{3 \times 4}\right) + ... + \left( \frac{ \sqrt{63} }{ \sqrt{62 \times 63} } - \frac{ \sqrt{62} }{ \sqrt{62 \times 63} } \right) + \left( \frac{ \sqrt{64} }{ \sqrt{63 \times 64} } - \frac{ \sqrt{63} }{63 \times 64}\right)$

$= \left(1 - \frac{1}{ \sqrt{2} } \right) +  \left( \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{3} } \right) + \left( \frac{1}{ \sqrt{3} } - \frac{1}{ \sqrt{4} }\right) + ... + \left( \frac{1}{ \sqrt{62} } - \frac{1}{ \sqrt{63} } \right) + \left( \frac{1}{ \sqrt{63} } - \frac{1}{ \sqrt{64} } \right)$

$= 1 - \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } - \frac{1}{\sqrt{3}} + \frac{1}{ \sqrt{3} } - \frac{1}{\sqrt{4}} + ... + \frac{1}{ \sqrt{62} } - \frac{1}{\sqrt{63}} + \frac{1}{ \sqrt{63} } - \frac{1}{\sqrt{64}}$

$= 1 - \frac{1}{\sqrt{64}} = 1 - \frac{1}{8} = \frac{7}{8}$