👤
Chiruadela85
a fost răspuns

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

Răspuns :

GreenEyes71

Salut,

[tex]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.[/tex]

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

Ai înțeles rezolvarea ?

Green eyes.

Răspuns:

Explicație pas cu pas:

[tex]\bf 113^{113} = 113^{112+1} = 113^{112}\cdot 113^1=[/tex]

[tex]\bf 113^{112}\cdot 113^1=113^{112}\cdot\big(49+64\big)=[/tex]

[tex]\bf 113^{56\cdot2}\cdot\big(7^2+8^{2} \big)=\big(113^{56} \big)^2\cdot\big(7^2+8^{2} \big)=[/tex]

[tex]\bf \big(113^{56} \big)^2\cdot7^2+\big(113^{56} \big)^2\cdot 8^{2} =[/tex]

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

             

Alte intrebari