Question

Desenvolva os binômios, deixando os cálculos:

( x + 5) 3 =




( 2x + 4) 5 =




c)(3x + 1 ) 4 =





d)( 2x + 8) 4=




e) (6x + 2) 6= ​

Answer 1

Explicação passo-a-passo:

a)

$\sf (x+5)^3=x^3+3\cdot x^2\cdot5+3\cdot x\cdot5^2+5^3$

$\sf (x+5)^3=x^3+15x^2+3x\cdot25+125$

$\sf (x+5)^3=x^3+15x^2+75x+125$

b)

$\sf (2x+4)^5=\dbinom{5}{0}\cdot(2x)^5\cdot4^0+\dbinom{5}{1}\cdot(2x)^4\cdot4^1+\dbinom{5}{2}\cdot(2x)^3\cdot4^2+\dbinom{5}{3}\cdot(2x)^2\cdot4^3+\dbinom{5}{4}\cdot(2x)^1\cdot4^4+\dbinom{5}{5}\cdot(2x)^0\cdot4^5$

$\sf (2x+4)^5=32x^5+5\cdot16x^4\cdot4+10\cdot8x^3\cdot16+10\cdot4x^2\cdot64+5\cdot2x\cdot256+1024$

$\sf (2x+4)^5=32x^5+320x^4+1280x^3+2560x^2+2560x+1024$

c)

$\sf (3x+1)^4=\dbinom{4}{0}\cdot(3x)^4\cdot1^0+\dbinom{4}{1}\cdot(3x)^3\cdot1^1+\dbinom{4}{2}\cdot(3x)^2\cdot1^2+\dbinom{4}{3}\cdot(3x)^1\cdot1^3+\dbinom{4}{4}\cdot(3x)^0\cdot1^4$

$\sf (3x+1)^4=81x^4+4\cdot27x^3\cdot1+6\cdot9x^2\cdot1+4\cdot3x\cdot1+1$

$\sf (3x+1)^4=81x^4+108x^3+54x^2+12x+1$

d)

$\sf (2x+8)^4=\dbinom{4}{0}\cdot(2x)^4\cdot8^0+\dbinom{4}{1}\cdot(2x)^3\cdot8^1+\dbinom{4}{2}\cdot(2x)^2\cdot8^2+\dbinom{4}{3}\cdot(2x)^1\cdot8^3+\dbinom{4}{4}\cdot(2x)^0\cdot8^4$

$\sf (2x+8)^4=16x^4+4\cdot8x^3\cdot8+6\cdot4x^2\cdot64+4\cdot2x\cdot512+4096$

$\sf (2x+8)^4=16x^4+256x^3+1536x^2+4096x+4096$

e)

$\sf (6x+2)^6=\dbinom{6}{0}\cdot(6x)^6\cdot2^0+\dbinom{6}{1}\cdot(6x)^5\cdot2^1+\dbinom{6}{2}\cdot(6x)^4\cdot2^2+\dbinom{6}{3}\cdot(6x)^3\cdot2^3+\dbinom{6}{4}\cdot(6x)^2\cdot2^4+\dbinom{6}{5}\cdot(6x)^1\cdot2^5+\dbinom{6}{6}\cdot(6x)^0\cdot2^6$

$\sf (6x+2)^6=46656x^6+6\cdot7776x^5\cdot2+15\cdot1296x^4\cdot4+20\cdot216x^3\cdot8+15\cdot36x^2\cdot16+6\cdot6x\cdot32+64$

$\sf (6x+2)^6=46656x^6+93312x^5+77760x^4+34560x^3+8640x^2+1152x+64$