Question

Va rog, dau Coroana

Ff urgent

Answer 1

Răspuns: $\color{red}\LARGE \boxed{\bf a^{b+c} = 1}$

Explicație pas cu pas:

$\LARGE \bf a,b,c \in \mathbb{N^{\star}}$

$\large \bf \dfrac{a^{2}+a }{2} +\dfrac{b^{2}+b}{2}= \dfrac{c+3}{c+1}$

$\large \bf \dfrac{a\cdot (a+1)}{2} +\dfrac{b\cdot (b+1)}{2}= \dfrac{c+3}{c+1}$

a·(a + 1)   si   b·(b + 1)    numere consecutive

    ↓                  ↓

   par               par        

$\large \bf \dfrac{a\cdot (a+1)}{2} +\dfrac{b\cdot (b+1)}{2}\in \mathbb{N}\implies c+1\big|c+3 \implies c+1\big|2\implies$

$\large \bf c+1\in \{1,2\}\implies\large\boxed{\bf c\in \{0,1\} }$

$\large \bf c=1 \implies \dfrac{a\cdot (a+1)}{2} =\dfrac{b\cdot (b+1)}{2}= 1\implies$

$\large \bf a\cdot (a+1)=2 \implies \Large\boxed{\bf a = 1}$

$\large \bf b\cdot (b+1)=2 \implies \Large\boxed{\bf b = 1}$

$\Large \bf a^{b+c} = 1^{1+1}\implies \Large\boxed{\bf a^{b+c} = 1}$

Bafta multa!

P.S.: Te rog sa glisezi spre stânga pentru a vedea toata rezolvarea daca esti pe telefon

==pav38==