Question

Resolva os produtos notáveis abaixo:

a) (2x2 + 4)2

b) (3a3 - 2a)2

2)Resolva as equações abaixo:

a)5x2 - 20x = 0

b)x2 - 100 = 0

c)x2 - x - 12 = 0​

Answer 1

Resposta:

$(2x2 + 4)2\\\\=2\left(2\cdot \:2x+4\right)\\\\=2\cdot \:2x\cdot \:2+2\cdot \:4\\\\=2\cdot \:2\cdot \:2x+2\cdot \:4\\\\=8x+8\\\\\\\\ (3a3 - 2a)2\\\\=2\left(9a-2a\right)\\\\=7\cdot \:2a\\\\=14a$

$5x2 - 20x = 0\\\\10x-20x=0\\\\-10x=0\\\\\frac{-10x}{-10}=\frac{0}{-10}\\\\x=0\\\\\\\\\\\\x2-100=0\\\\x\cdot \:2-100+100=0+100\\\\x\cdot \:2=100\\\\\frac{x\cdot \:2}{2}=\frac{100}{2}\\\\x=50\\\\\\\\\\x2 - x - 12 = 0\\\\x-12=0\\\\x-12+12=0+12\\\\x=12$

Answer 2

Explicação passo-a-passo:

a) (2x²+4)²= (2x²)²+2.2.4x²+4²= 4x⁴+16x²+16

b) (3a³-2a)²= (3a³)²-2.3.2a³a+(-2a)²= 9a⁶-12a⁴+4a²

2) 5x²-20x=0

a) 5x.(x-20)=0

x=0

x-20=0

x=20

b)x²-100=0

x²=100

$x = + ou - \sqrt{100} = + 10 \: \: e \: \: - 10$

c)x²-x-12=0

a=1, b= -1, e c=-12

delta =b²-4ac

delta =(-1)²-4x1x-12

delta =1+48

delta =49

$\sqrt{49} = 7$

x'= [-(-1)+7)]/2.1=

x'=[1+7]/2=

x'=8/2

x'=4

x"=[-(-1)-7]/2.1

x"=[1-7]/2

x"=-6/2

x"=-3