Question

Sa ma ajute cineva va rog. apasati pe imagine sa se vada tot. am nevoie la toate. repede daca se poate​

Answer 1

a)

$\bf 5^{1256}+5^{1257}+3\cdot5^{1255}=$

$\bf 5^{1255}\cdot(5^{1}+5^{2}+3\cdot5^{0})=$

$\bf 5^{1255}\cdot(5^{1}+25+3)=$

$\boxed{\bf 5^{1255}\cdot 33}$

b)

$\bf 2^{102}-2^{101}-2^{100}=$

$\bf 2^{100}\cdot(2^{2}-2^{1}-2^{0}) =$

$\bf 2^{100}\cdot(4-2-1) =$

$\bf 2^{100}\cdot 1 = \boxed{\bf 2^{100}}$

c)

$\bf (3^{75}-3^{74}-3^{73}):(5\cdot 3^{72})=$

$\bf 3^{73}(3^{2}-3^{1}-3^{0}):(5 \cdot 3^{72})=$

$\bf 3^{73}(9-3-1):(5 \cdot 3^{72})=$

$\bf 3^{73}\cdot 5 : 5 \cdot 3^{72}=$

$\bf 3^{73-72}=3^{1}=\boxed{\bf 3}$

d)

$\bf 5^{2n+1}-4\cdot 25^{n}- 5^{2n}=$

$\bf 5^{2n+1}-4\cdot 5^{2n}- 5^{2n}=$

$\bf 5^{2n}\cdot(5^{1}-4\cdot 5^{0}- 5^{0})=$

$\bf 5^{2n}\cdot(5-4-1)=$

$\bf 5^{2n}\cdot 0 =\boxed{\bf 0}$

e)

$\bf (7^{100}-6\cdot7^{99}+11^{110}: 121^{55}-7^{99})^{100}=$

$\bf (7^{100}-6\cdot7^{99}+11^{110}: 11^{110}-7^{99})^{100}=$

$\bf (7^{100}-6\cdot7^{99}+1-7^{99})^{100}=$

$\bf (7^{100}-7\cdot7^{99}+1)^{100}=$

$\bf (7^{100}-7^{100}+1)^{100}=$

$\bf 1^{100}=\boxed{\bf 1}$

f)

$\bf 2^{n+2}\cdot3^{n+1}-6^{n}=$

$\bf 2^{n+2}\cdot3^{n+1}-2^{n}\cdot3^{n}=$

$\bf 2^{n}\cdot3^{n}\cdot(2^{2}\cdot3^{1}-2^{0}\cdot3^{0})=$

$\bf 2^{n}\cdot3^{n}\cdot(4\cdot3-1\cdot1)=$

$\bf 2^{n}\cdot3^{n}\cdot(12-1)=$

$\boxed{\bf 2^{n}\cdot3^{n}\cdot 11}$

Answer 2

Explicație pas cu pas:

Am atasat rezolvarea ... .....