Question

4. Următoarele numere sunt pătrate perfecte? a) 1+3+5+9 b) 1+3+5+ 17 c) (2.1+1)+(2.2+1)+(2·3+1)+...+(2-16+1) d) 1+3+5+...+101 e) 5+5.4 g) 3-2-3-2 i) 34 k) 24.3² m) 17+8 locuiest casetele cu numere naturale ne- f) 129 + 129 128 h) 209-319-209-319 j) 514 1) 26-38.510


va rog rpd!!!​

Answer 1

Răspuns:

Explicație pas cu pas:

a) 1+3+5+9 = 18 = 2×3² → nu este pătrat perfect

b) 1+3+5+17 = 26 = 2×13→ nu este pătrat perfect

c) (2×1+1)+(2×2+1)+.....+(2×16+1) =

= 3+5+7+......+33 =

(33-3):2+1=30:2+1=16 termeni are suma

= 16×(3+33) : 2 =

= 8× 36 =

= 2³×2²×3² =

= 2⁵x3²→ nu este pătrat perfect

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d)  1+3+5+......+101 =

(101-1):2+1 = 51 termeni are suma ( primele 51 de numere impare)

= 51 × (1+101) : 2 =

= 51 × 102 : 2 =

= 51 × 51 =

= 51² → pătrat perfect

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e) 5 + 5× 4 = 5×(1+4) = 5×5 = 5² → pătrat perfect

  → l-am dat factor comun pe 5

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f) 129 + 129 ×128 = 129 × (1+128) = 129 × 129 = 129² → pătrat perfect

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g)  3×2×3×2 = 3²×2² = (3×2)² = 6² → pătrat perfect

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h)  209 ×319 × 209×319 = 209²×319² = (209×319)² → pătrat perfect

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i) 3⁴ = 3²ˣ² = (3²)² = 9² → pătrat perfect

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j)  5¹⁴ = 5⁷ˣ² = (5⁷)² → pătrat perfect

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k)  2⁴ x 3² = (2²)² x3² = (4x3)² = 12² → pătrat perfect

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l)   2⁶ x 3⁸ x 5¹⁰ = 2⁽³ˣ²⁾x 3⁽⁴ˣ²⁾ x 5⁽⁵ˣ²⁾ = (2³)² x (3⁴)²x(5⁵)²

= (8x3⁴x5⁵)² → pătrat perfect

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m)   17+8 = 25 = 5² → pătrat perfect