Question

Dau 30 puncte + coroana.

Answer 1

[tex]a)\\(x-3)(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2014})=0\\ \frac{1}{1}+\frac{1}{2}+...+\frac{1}{2014} este \ diferit \ de \ 0\rightarrow x-3=0\\x=3[/tex][tex]b)\\ (x-2)(\frac {1}{1\cdot 2}+\frac {1}{2\cdot 3}+....\frac {1}{2014\cdot 2015})=2014\\ (x-2)(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...\frac{2015-2014}{2014\cdot 2015}=2014\\ (x-2)(\frac{ 2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+....+\frac{2015}{2014\cdot2015}-\frac{2014}{2014\cdot2015})=2014\\ (x-2)(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2014}-\frac{1}{2015})=2014\\ (x-2)(1-\frac{1}{2015})=2014\\ (x-2)\cdot\frac{2014}{2015}=2014 |:2014\\ \frac{x-2}{2015}=1[/tex][
$x-2=2015\\x=2017$

Answer 2

[tex]\it a)\ \dfrac{x-3}{1} +\dfrac{x-3}{2} +\dfrac{x-3}{3}+\ ...\ +\dfrac{x-3}{2004} =0\ \Leftrightarrow \\\;\\ \\\;\\ \Leftrightarrow (x-3)\left(\dfrac{1}{1} +\dfrac{1}{2} +\dfrac{1}{3}+\ ...\ +\dfrac{1}{2004}\right) =0\ \Leftrightarrow x-3=0 \Leftrightarrow x=3[/tex]



[tex]\it\ b)\ \dfrac{x-2}{2}+\dfrac{x-2}{6}+\dfrac{x-2}{12}+\ ...\ +\dfrac{x-2}{2014\cdot2015} =2014 \ \Leftrightarrow \\\;\\ \\\;\\ \Leftrightarrow (x-2)\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\ ...\ +\dfrac{1}{2014\cdot2015}\right)=2014\ \Leftrightarrow \\\;\\ \\\;\\ \Leftrightarrow \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\ ...\ +\dfrac{1}{2014\cdot2015} =\dfrac{2014}{x-2}[/tex]

Pentru membrul stâng al ultimei egalități, folosim descompunerea :

$\it \dfrac{1}{k(k+1)} =\dfrac{1}{k}-\dfrac{1}{k+1}$

Prin urmare, vom avea:

[tex]\it 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ ...\ +\dfrac{1}{2014}-\dfrac{1}{2015} =\dfrac{2014}{x-2} \Longrightarrow \\\;\\ \\\;\\ \Longrightarrow 1-\dfrac{1}{2015}=\dfrac{2014}{x-2} \Longrightarrow \dfrac{2014}{2015} = \dfrac{2014}{x-2} \Longrightarrow x-2=2015 \Longrightarrow \\\;\\ \\\;\\ \Longrightarrow x=2015+2 \Longrightarrow x=2017[/tex]