Răspuns :

Florin3364id

Răspuns:

Explicație pas cu pas:

observam mai intai ca:

[tex]\frac{1}{x*(x+1)} = \frac{(x+1) - x}{x*(x+1)} = \frac{x+1}{x*(x+1)} -\frac{x}{x*(x+1)} = \frac{1}{x} -\frac{1}{x+1}[/tex]

deci

[tex]a = \sqrt{\frac{1}{2}*(\frac{1}{1*2} +\frac{1}{2*3} +\frac{1}{3*4} + ... +\frac{1}{48*49} +\frac{1}{49*50}) } =[/tex]

[tex]= \sqrt{\frac{1}{2}*[(\frac{1}{1} - \frac{1}{2}) +(\frac{1}{2} - \frac{1}{3}) +(\frac{1}{3} - \frac{1}{4}) + ... +(\frac{1}{48} - \frac{1}{19}) +(\frac{1}{49} - \frac{1}{50}) }] } =[/tex]

[tex]= \sqrt{\frac{1}{2}*(\frac{1}{1} - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4} + ... +\frac{1}{48} - \frac{1}{19} +\frac{1}{49} - \frac{1}{50} }) } =[/tex]

[tex]= \sqrt{\frac{1}{2}*(1 - \frac{1}{50} }) } = \sqrt{\frac{1}{2}*\frac{50-1}{50} } } = \sqrt{\frac{49}{100} } = \sqrt{\frac{7^2}{10^2} }= \sqrt{(\frac{7}{10})^2 } = \frac{7}{10}[/tex]

Răspuns: [tex]\red{\boxed{\bf ~a = \dfrac{7}{10}~}}[/tex]

Explicație pas cu pas:

Ne folosim de regulile sumelor telescopice

[tex]\bf a= \sqrt{\dfrac{1}{2}\cdot\bigg[ \bigg(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{48\cdot49}+\dfrac{1}{49\cdot50}\bigg)\bigg]}[/tex]

[tex]a= \sqrt{\dfrac{1}{2}\cdot\bigg[ \bigg(\dfrac{1}{1}-\dfrac{1}{2}\bigg)+\bigg(\dfrac{1}{2}-\dfrac{1}{3}\bigg)+\bigg(\dfrac{1}{3}-\dfrac{1}{4}\bigg)+...+\bigg(\dfrac{1}{48}-\dfrac{1}{49}\bigg)+\bigg(\dfrac{1}{49}-\dfrac{1}{50}\bigg)\bigg]}[/tex]

[tex]a= \sqrt{\dfrac{1}{2}\cdot \bigg(\dfrac{1}{1}-\not\dfrac{1}{2}+\not\dfrac{1}{2}-\not\dfrac{1}{3}+\not\dfrac{1}{3}-\not\dfrac{1}{4}+\not~....\not ~+\not\dfrac{1}{48}-\not\dfrac{1}{49}+\not\dfrac{1}{49}-\dfrac{1}{50}\bigg)}[/tex]

[tex]a= \sqrt{\dfrac{1}{2}\cdot \bigg(\dfrac{^{50)} 1~}{1}-\dfrac{1}{50}\bigg)}= \sqrt{\dfrac{1}{2}\cdot \bigg(\dfrac{50}{50}-\dfrac{1}{50}\bigg)}[/tex]

[tex]a= \sqrt{\dfrac{1}{2}\cdot \bigg(\dfrac{50-1}{50}\bigg)}= \sqrt{\dfrac{1}{2}\cdot \dfrac{49}{50}}=\sqrt{\dfrac{49}{100}}[/tex]

[tex]a= \sqrt{\dfrac{7^{2} }{10^{2}}}= \sqrt{\bigg(\dfrac{7}{10}\bigg)^{2}}\implies\red{\boxed{\bf ~a=\dfrac{7}{10}~}}[/tex]

P.S.:

Dacă ești pe telefon te rog să glisezi spre stânga pentru a vedea întreaga rezolvarea (benefic este să vezi rezolvarea pe laptop sau PC). Am încercat să fiu cât mai succint dar și explicit în același timp.

Baftă multă !

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