Question

Sa se rezolve urmatoarea problema

Answer 1

 

$\displaystyle\bf\\13^x+16^x=15^x+14^x\\\\\text{Observam ca suma bazelor puterilor este}~~13 + 16 = 15+14=29\\\text{Rezulta ca }~x=1~\text{ verifica ecuatia.}\\\text{Verificare}\\13^1+16^1=15^1+14^1\\29 = 29\\\implies~~\boxed{\bf x_1=1}\\\\----------\\\text{Avem 2 puteri in stanga egalului si 2 puteri in dreapta.}\\\text{Rezulta ca }~x=0~\text{ verifica ecuatia.}\\\\\text{Verificare}\\13^0+16^0=15^0+14^0\\1+1 = 1+1\\2=2\\\implies~~\boxed{\bf x_2=0}$

Answer 2

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.

$13^x+16^x=15^x+14^x\\16^x-15^x=14^x-13^2\\\dfrac{16^x-15^x}{16-15}=\dfrac{14^x-13^x}{14-13}\\\texttt{Consideram functiile f,g astfel incat }f:[15,16]\to\mathbb{R},\\g:[13,14]\to\mathbb{R}, f(t)=g(t)=t^x.\texttt{ Deoarece f este continua pe }[15,16]\\\texttt{si derivabila pe (15,16), conform teoremei lui Lagrange }\\\exists~c\in[15,16]\texttt{ astfel incat }f'(c)=\dfrac{f(16)-f(15)}{16-15}, \texttt{de unde rezulta }\\x\cdot c^{x-1}=\dfrac{16^x-15^x}{16-15}$

$\texttt{Analog,}\exists~d\in[13,14]\texttt{ astfel incat }g'(d)=\dfrac{g(14)-g(13)}{14-13},\texttt{de unde}\\x\cdot d^{x-1}=\dfrac{14^x-13^x}{14-13}.\texttt{ Egalitatea devine:}\\x\cdot c^{x-1}=x\cdot d^{x-1}\\x\cdot(c^{x-1}-d^{x-1})=0\\i)x=0\texttt{ este solutie }\\ii)c^{x-1}-d^{x-1}=0\\c^{x-1}=d^{x-1}\\\left(\dfrac{c}{d}\right)^{x-1}=1\\x-1=0\Rightarrow x=1\\\texttt{Prin urmare, singurele solutii convenabile sunt 0 si 1.}$