Question

Scrieți următoarele Nr sub forma de produs utilizând factorul comun
A)2^76+2^78+2^80
B)3*5^47+2*5^48+6*5^49
C)13^14 *3-13^13*2-13^12
Daca vad. Explicați dau coroniță

Answer 1

Răspuns:

$\bf a) ~2^{76} + 2^{78} + 2^{80} =$

$\bf 2^{76}\cdot \big(2^{76-76} + 2^{78-76} + 2^{80-76}\big)=$

$\bf 2^{76}\cdot \big(2^{0} + 2^{2} + 2^{4}\big)=2^{76}\cdot \big(1 + 4 +16\big)=$

$\blue{\underline{~\bf 2^{76}\cdot21~}}$

                               

$\bf b) ~3\cdot5^{47} + 2\cdot 5^{48} + 6\cdot5^{49} =$

$\bf 5^{47}\cdot \big(3\cdot5^{47-47} + 2\cdot5^{48-47} +6\cdot5^{49-47}\big)=$

$\bf 5^{47}\cdot \big(3\cdot5^{0} + 2\cdot5^{1} +6\cdot5^{2}\big)=$

$\bf 5^{47}\cdot \big(3\cdot 1 + 2\cdot5 +6\cdot 25\big)= 5^{47}\cdot \big(3+10+150\big)=$

$\purple{\underline{~\bf 5^{47}\cdot163~}}$

                               

$\bf c) ~13^{14}\cdot 3 -13^{13}\cdot2- 13^{12} =$

$\bf 13^{12}\cdot \big(13^{14-12}\cdot 3 -13^{13-12}\cdot2- 13^{12-12}\big)=$

$\bf 13^{12}\cdot \big(13^{2}\cdot 3 -13^{1}\cdot2- 13^{0}\big)=$

$\bf 13^{12}\cdot \big(169\cdot 3 -13\cdot2- 1\big)=$

$\bf 13^{12}\cdot \big(507 -26- 1\big)=$

$\red{\underline{~\bf 13^{12}\cdot 480~}}$