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Sorinaleach
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Aratati ca nr a este patrat perfect: a=1+2+3+...+264+265•133

Răspuns :

Albastruverde12
[tex]A=1+2+3+...+264 +265 \cdot 133= \\ \\ = \frac{264 \cdot 265}{2}+265 \cdot 133= \\ \\ =132 \cdot 265 +265 \cdot 133= \\ \\ =265(132+133)= \\ \\ =265 \cdot 265= \\ \\ =265 ^2 -patrat~perfect [/tex]
Max13
[tex]A=1+2+...+264+265*133= \frac{264(264+1)}{2}+265*133=132*265+ [/tex][tex]265*133=265(133+132)= 265^{2} [/tex]-pătrat perfect.

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