Question

3 resolva os produtos notáveis abaixo :
a) (2×+1)^2
b) (2×+3)^2
c) (×+1/3)^2
d) (2×-5)^2
e) (2×-7)^2
f) (×-1/4)^2
g) (×-2)(×+2)
h)(a-2b)(a+2b)
i) (2×+y)(2×-y)
j)(x+1/2)(×-1/2)


por favor é pra agora ​

Answer 1

Explicação passo-a-passo:

a)

(2×+1)² = [ ( 2x)² + 2 * 2x * 1 + ( 1)² ] = 4x² + 4x + 1 >>>>

b)

(2×+3)² = [ ( 2x)² + 2 * 2x * 3 + ( 3 )² ] = 4x² + 12x + 9 >>>>

c)

(×+1/3)² = [ ( x )² + 2* x * 1/3 + ( 1/3)² ] = x² + 2x/3 + 1/9 >>>>

d)

( 2x - 5)² = [ ( 2x )² - 2 * 2x * 5 + ( 5 )² ] = 4x² - 20x + 25 >>>>

e)

(2x - 7 )²= [ ( 2x)² - 2 * 2x * 7 + ( 7)² ] = 4x² - 28x + 49 >>>

f)

( x - 1/4 )² =[ ( x )² - 2 * x * 1/4 + ( 1/4)² ] = x² - 2x/4 + 1/16 ]

Notas

Simplificando 2x/4 por 2 = 1x/2 >>>

( 1/4)² = 1/4 * 1/4 = 1/16 >>>

reescrevendo

x² - 1x/2 + 1/16 >>>>>resposta

g)

(x-2)(x+2) = soma pela diferença = [ ( x)² - ( 2)² ] = x² - 4 >>>

h)

(a-2b)(a+2b)=idem = [ ( a )² - ( 2b)² ] = a² - 4b² >>>>

i)

( 2x +y)(2x -y) = idem = { ( 2x)² - ( y )² ] =4x² - y² >>>>

j)

(x+1/2)(x -1/2) =idem [ ( x )² - ( 1/2)² ] = x² - 1/4 >>>>

Nota >>. 1/2 * 1/2 = 1/4 >>>