Răspuns:
Explicație pas cu pas:
Na fii atent acilea la demonstratia aia blanao .
Dupa cum ziceam voi folosi teorema lui Lagrange, pana una alta fac un artificiu de calcul.
[tex]13^x+16^x=15^x+14^x\\16^x-15^x=14^x-13^2\\\dfrac{16^x-15^x}{16-15}=\dfrac{14^x-13^x}{14-13}\\\texttt{Consideram functiile f,g astfel incat }f:[15,16]\to\mathbb{R},\\g:[13,14]\to\mathbb{R}, f(t)=g(t)=t^x.\texttt{ Deoarece f este continua pe }[15,16]\\\texttt{si derivabila pe (15,16), conform teoremei lui Lagrange }\\\exists~c\in[15,16]\texttt{ astfel incat }f'(c)=\dfrac{f(16)-f(15)}{16-15}, \texttt{de unde rezulta }\\x\cdot c^{x-1}=\dfrac{16^x-15^x}{16-15}[/tex]
[tex]\texttt{Analog,}\exists~d\in[13,14]\texttt{ astfel incat }g'(d)=\dfrac{g(14)-g(13)}{14-13},\texttt{de unde}\\x\cdot d^{x-1}=\dfrac{14^x-13^x}{14-13}.\texttt{ Egalitatea devine:}\\x\cdot c^{x-1}=x\cdot d^{x-1}\\x\cdot(c^{x-1}-d^{x-1})=0\\i)x=0\texttt{ este solutie }\\ii)c^{x-1}-d^{x-1}=0\\c^{x-1}=d^{x-1}\\\left(\dfrac{c}{d}\right)^{x-1}=1\\x-1=0\Rightarrow x=1\\\texttt{Prin urmare, singurele solutii convenabile sunt 0 si 1.}[/tex]
