Question

AM NEVOIE URGENT!!!! DAU COROANĂ!!!​

Answer 1

Cred ca e bine.Eu am făcut asa

Answer 2

Răspuns:

Explicație pas cu pas:

$\frac{1}{1\cdot2} + \frac{1}{2\cdot 3} + ... + \frac{1}{24\cdot 25} = \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{24} - \frac{1}{25} \\ = 1 - \frac{1}{25} = \frac{25 - 1}{25} = \frac{24}{25}$

$a=\sqrt{2(\frac{24*25}{2} (\frac{24}{25} )})=\sqrt{24^{2} }=24$

$\frac{1}{1\cdot2} + \frac{1}{2\cdot 3} + ... + \frac{1}{81\cdot 82} = \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{81} - \frac{1}{82} \\ = 1 - \frac{1}{82} = \frac{82 - 1}{82} = \frac{81}{82}$

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

$c=\frac{\sqrt{2} - \sqrt{1}}{ \sqrt{1 \cdot 2}} + \frac{\sqrt{3} - \sqrt{2}}{\sqrt{2 \cdot 3}} + \frac{\sqrt{4} - \sqrt{3}}{\sqrt{3 \cdot 4}} + ... + \frac{\sqrt{64} - \sqrt{63}}{\sqrt{64 \cdot 63}} = \frac{\sqrt{2}}{ \sqrt{2} } - \frac{1}{\sqrt{2}} + \frac{1}{ \sqrt{2} } - \frac{1}{\sqrt{3}} + \frac{1}{ \sqrt{3} } - \frac{1}{\sqrt{4}} + ... + \frac{1}{ \sqrt{63} } - \frac{1}{\sqrt{64}} = 1 - \frac{1}{\sqrt{64}}=1-\frac{1}{8}=\frac{7}{8}$