Question

scrie numarul 113 la puterea 113 ca suma de 2 patrate perfecte.​

Answer 1

Salut,

$113^{113}=113^{1+112}=113\cdot 113^{112}=(49+64)\cdot 113^{56\cdot 2}=(7^2+8^2)\cdot (113^{56})^2=\\\\=7^2\cdot (113^{56})^2+8^2\cdot (113^{56})^2=(7\cdot 113^{56})^2+(8\cdot 113^{56})^2.$

Am obținut o sumă de 2 pătrate perfecte, ceea ce trebuia demonstrat.

Ai înțeles rezolvarea ?

Green eyes.

Answer 2

Răspuns:

Explicație pas cu pas:

$\bf 113^{113} = 113^{112+1} = 113^{112}\cdot 113^1=$

$\bf 113^{112}\cdot 113^1=113^{112}\cdot\big(49+64\big)=$

$\bf 113^{56\cdot2}\cdot\big(7^2+8^{2} \big)=\big(113^{56} \big)^2\cdot\big(7^2+8^{2} \big)=$

$\bf \big(113^{56} \big)^2\cdot7^2+\big(113^{56} \big)^2\cdot 8^{2} =$

$\red{\boxed{\bf\Big(113^{56}\cdot 7\Big)^2+\Big(113^{56} \cdot 8\Big)^{2}\Rightarrow suma~ de~ doua~ patrate~ perfecte}}$