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»PROBLEME DIVERSE CU REZOLVARI COMPLETE-LICEU

Data publicarii: 11 Octombrie, 2010

PROBLEMA-25

Suport teoretic:

Partea intreaga a unui numar real, identitatea lui Hermite, sisteme de inecuatii, conditii de existenta.

Enunt:

Sa se rezolve in multimea numerelor reale urmatoarea ecuatie, unde [a] reprezinta

partea intreaga a numarului real a:

$\frac{1}{[x]}+\frac{1}{[x+\frac{1}{2}]}=[2x].$ \frac{1}{[x]}+\frac{1}{[x+\frac{1}{2}]}=[2x].

Raspuns:

${x}\in{[-1,-\frac{1}{2})}\cup{[1,\frac{3}{2})}.$ {x}\in{[-1,-\frac{1}{2})}\cup{[1,\frac{3}{2})}.

Rezolvare:

Ecuatia se mai poate scrie sub forma:

$\frac{[x]+[x+\frac{1}{2}]}{{[x]}\cdot{[x+\frac{1}{2}]}}=[2x].$ \frac{[x]+[x+\frac{1}{2}]}{{[x]}\cdot{[x+\frac{1}{2}]}}=[2x].

Sa observam mai intai ca prezenta numitorilor impune conditiile de existenta:

$[x]\not={0}\;si\;[x+\frac{1}{2}]\not={0},\;sau:$ [x]\not={0}\;si\;[x+\frac{1}{2}]\not={0},\;sau:

$\begin{cases}{x}\in{\mathbb{R}}\setminus{[0,1)}\\{x}\in{\mathbb{R}}\setminus{[-\frac{1}{2},\frac{1}{2})}\end{cases},$ \begin{cases}{x}\in{\mathbb{R}}\setminus{[0,1)}\\{x}\in{\mathbb{R}}\setminus{[-\frac{1}{2},\frac{1}{2})}\end{cases},

de unde rezulta ca

${x}\in{(-\infty,-\frac{1}{2})}\cup{[1,+\infty)}.\;(1)$ {x}\in{(-\infty,-\frac{1}{2})}\cup{[1,+\infty)}.\;(1)

Tinand cont de identitatea lui Hermite (cazul n=2), anume

$[x]+[x+\frac{1}{2}]=[2x],\;\forall{x}\in{\mathbb{R}},$ [x]+[x+\frac{1}{2}]=[2x],\;\forall{x}\in{\mathbb{R}},

ecuatia devine:

$\frac{[2x]}{[x]\cdot[x+\frac{1}{2}]}=[2x].$ \frac{[2x]}{[x]\cdot[x+\frac{1}{2}]}=[2x].

De aici rezulta

$[2x]=0$ [2x]=0 $\Leftrightarrow$ \Leftrightarrow ${0}\le{2x}<{0}$ {0}\le{2x}<{0} $\Leftrightarrow$ \Leftrightarrow ${x}\in{[0,\frac{1}{2})},\;(2)$ {x}\in{[0,\frac{1}{2})},\;(2)

$[x]\cdot[x+\frac{1}{2}]=1.$ [x]\cdot[x+\frac{1}{2}]=1.

Prin definitie, partea intreaga a unui numar real este un numar intreg, deci distingem

cazurile:

$I)\begin{cases}{[x]=1}\\{[x+\frac{1}{2}]=1}\end{cases}$ I)\begin{cases}{[x]=1}\\{[x+\frac{1}{2}]=1}\end{cases} $\Leftrightarrow$ \Leftrightarrow $\begin{cases}{1}\le{x}<{2}\\{1}\le{x+\frac{1}{2}}<{2}\end{cases},$ \begin{cases}{1}\le{x}<{2}\\{1}\le{x+\frac{1}{2}}<{2}\end{cases},

de unde obtinem:

${x}\in{[1,\frac{3}{2})}.\;(3)$ {x}\in{[1,\frac{3}{2})}.\;(3)

$II)\begin{cases}{[x]=-1}\\{[x+\frac{1}{2}]=-1}\end{cases}$ II)\begin{cases}{[x]=-1}\\{[x+\frac{1}{2}]=-1}\end{cases} $\Leftrightarrow$ \Leftrightarrow $\begin{cases}{-1}\le{x}<{0}\\{-1}\le{x+\frac{1}{2}}<{0}\end{cases},$ \begin{cases}{-1}\le{x}<{0}\\{-1}\le{x+\frac{1}{2}}<{0}\end{cases},