utilizarea factorului comun 7 × 4²³-9×4²²=19×4²² ,49¹¹:7¹¹+7¹³:7=8¹¹, 25¹⁰ × 5²³×125⁵,9⁴×27⁵×3¹⁴, (a×b)¹²:a⁴×b³×(a2×b), 36⁵×6⁵×6×+¹
Răspunsuri la întrebare
Răspuns:
7 × 4²³- 9 × 4²²= 19 × 4²²
4²² × (7 × 4²³⁻²² - 9 × 4²²⁻²²) =
4²² × (7 × 4¹ - 9 × 4⁰) =
4²² × (28 - 9 × 1) =
4²² × (28 - 9) = 4²² × 19
49¹¹ : 7¹¹ + 7¹³ : 7 = 8¹¹
(7²)¹¹ : 7¹¹ + 7¹³⁻¹ =
7²ˣ¹¹ : 7¹¹ + 7¹² =
7²² : 7¹¹ + 7¹² =
7²²⁻¹¹ + 7¹² = 7¹¹ + 7¹² =
7¹¹ × (7¹¹⁻¹¹ + 7¹²⁻¹¹) =
7¹¹ × (7⁰ + 7¹) =
7¹¹ × (1 + 7) = 7¹¹ × 8
25¹⁰ × 5²³ × 125⁵ =
(5²)¹⁰ × 5²³ × (5³)⁵ =
5²⁰ × 5²³ × 5¹⁵ =
5²⁰⁺²³⁺¹⁵ = 5⁵⁸
9⁴ × 27⁵ × 3¹⁴ =
(3²)⁴ × (3³)⁵ × 3¹⁴ =
3²ˣ⁴ × 3³ˣ⁵ ×3¹⁴ =
3⁸ × 3¹⁵ ×3¹⁴ =
3⁸⁺¹⁵⁺¹⁴ = 3³⁷
(a × b)¹² : a⁴ × b³ × (a² × b) =
a¹² × b¹² : a⁴ × b³ × a² × b =
a¹²⁻⁴⁺² × b¹²⁺³⁺¹ = a¹⁰ × b¹⁵
36⁵ × 6⁵ × 6ˣ⁺¹ =
(6²)⁵ × 6⁵ × 6ˣ⁺¹ =
6²ˣ⁵ × 6⁵ × 6ˣ⁺¹ =
6¹⁰ × 6⁵ × 6ˣ⁺¹ =
6¹⁰⁺⁵ × 6ˣ⁺¹ = 6¹⁵ × 6ˣ⁺¹ =
6¹⁵⁺ˣ⁺¹ = 6ˣ+¹⁶
==pav38==
Răspuns:
7 × 4²³-9×4²²=4²²(7·4-9·1)=4²²(28-9)=4²²·19
49¹¹:7¹¹+7¹³:7=(7²)¹¹:7¹¹+7¹²=7²²⁻¹¹+7¹²=7¹¹(1+7)=7¹¹·8
25¹⁰ × 5²³×125⁵=(5²)¹⁰5²²(5³)⁵=5²⁰·5²²·5¹⁵=5²⁰⁺²²⁺¹⁵=5⁵⁸
9⁴×27⁵×3¹⁴= (3²)⁴·(3³)⁵· 3¹⁴= 3⁸·3¹⁵·3¹⁴=3⁸⁺¹⁵⁺¹⁴=3³⁷
(a×b)¹²:a⁴×b³×(a²×b)= a¹²·b¹²:a⁴·b³·a²·b = a¹²⁻⁴⁺²·b¹²⁺³⁺¹=a¹⁰·b¹⁵
36⁵×6⁵×6×⁺¹=(6²)⁵·6⁵·6ˣ⁺¹=6¹⁰⁺⁵⁺ˣ⁺¹=6ˣ⁺¹⁶