Sa se rezolve ecuatia:
\( x\cdot 2^{x-1}+\frac{1}{\sqrt{x}}\cdot 2^{\frac{1}{\sqrt{x}}=3, x>0 \).
Cezar Lupu, I.V. Maftei, R.M.I. C-ta 2005
Ecuatie exponentiala si radicali
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Ecuatie exponentiala si radicali
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
Din inegalitatea lui Cebisev avem ca :
\( x2^{x-1}+\frac{1}{\sqrt x}2^{\frac{1}{\sqrt x}}=x2^{x-1}+\frac{1}{ \sqrt x}2^{\frac{1}{\sqrt x}-1}+\frac{1}{ \sqrt x}2^{\frac{1}{\sqrt x} -1}\geq \) \( \geq \frac{1}{3} \left( x+\frac{1}{ \sqrt x }+\frac{1}{ \sqrt x } \right) \left( 2^{x-1}+2^{\frac{1}{ \sqrt x}-1}+2^{\frac{1}{ \sqrt x}-1} \right) \)
Din \( x+\frac{1}{ \sqrt x }+\frac{1}{ \sqrt x }\geq 3 \) (medii) si \( 2^{x-1}+2^{\frac{1}{ \sqrt x}-1}+2^{\frac{1}{ \sqrt x}-1} \geq 3 \sqrt[3]{2^{x+\frac{1}{\sqrt x}+\frac{1}{\sqrt x}-3}}\geq 3 \) obtinem ca \( x2^{x-1}+\frac{1}{\sqrt x}2^{\frac{1}{\sqrt x}}\geq 3 \)
Cum avem egalitate in inegalitate avem ca \( x=1 \) este unica solutie, cum \( x=1 \)este conditia de egalitate
\( x2^{x-1}+\frac{1}{\sqrt x}2^{\frac{1}{\sqrt x}}=x2^{x-1}+\frac{1}{ \sqrt x}2^{\frac{1}{\sqrt x}-1}+\frac{1}{ \sqrt x}2^{\frac{1}{\sqrt x} -1}\geq \) \( \geq \frac{1}{3} \left( x+\frac{1}{ \sqrt x }+\frac{1}{ \sqrt x } \right) \left( 2^{x-1}+2^{\frac{1}{ \sqrt x}-1}+2^{\frac{1}{ \sqrt x}-1} \right) \)
Din \( x+\frac{1}{ \sqrt x }+\frac{1}{ \sqrt x }\geq 3 \) (medii) si \( 2^{x-1}+2^{\frac{1}{ \sqrt x}-1}+2^{\frac{1}{ \sqrt x}-1} \geq 3 \sqrt[3]{2^{x+\frac{1}{\sqrt x}+\frac{1}{\sqrt x}-3}}\geq 3 \) obtinem ca \( x2^{x-1}+\frac{1}{\sqrt x}2^{\frac{1}{\sqrt x}}\geq 3 \)
Cum avem egalitate in inegalitate avem ca \( x=1 \) este unica solutie, cum \( x=1 \)este conditia de egalitate
Vrajitoarea Andrei