Question

01) Calcule
a) (3 + x)2 =
b) (x + 5)2 =
c) (x + y)2 =
d) (x + 2)2 =
e) ( 3x + 2)2 =
f) (2x + 1)2 =
g) (a? + 1)2 =
h) (y3 + 3)2 =
i) (a + b)2 =
j) (x + 2y3)2 =​

Answer 1

Explicação passo a passo:

QUADRADO DA SOMA > regra > QUADRADO DO PRIMEIRO TERMO,MAIS 2 VEZES OPRIMEIRO VEZES O SEGUNDO, MAIS O QUADRADO DO SEGUNDO

a)

(3 + x)² =[ ( 3)² + 2* 3 * X + (X)² ] = 9 + 6X + X² >>>resposta

b)

(x + 5)² = [ ( X )² +2 * 5 * X +( 5)²] = X² + 10X+ 25>>>>resposta

c)

(x + y)² = [ (X)² + 2 * X * Y + ( Y)² ] = X² + 2XY + Y² >>>>resposta

d)

(x + 2)² = [ ( X)² + 2 * X * 2 + ( 2 )² ] = X² + 4X + 4 >>>>resposta

e)

( 3x + 2)² = [ (3X)² + 2 * 3X * 2 + ( 2)² ] = 9X² + 12X + 4 >>>>resposta

f)

(2x + 1)² = [(2X)² + 2 * 2X * 1 + ( 1 )² ] = 4X² + 4X + 1 >>>>>resposta

g) (a? + 1)² = INCOMPLETA

h)

(y³ + 3)² = [ ( y³)² + 2 * y³ * 3 = ( 3)² ] = y^6 + 6y³ +9 >>>>resposta

i)

(a + b)² = [ ( a)² + 2 * a * b + ( b)²] = a² + 2ab + b² >>>>resposta

j)

(x + 2y³)² =​[ ( x )² + 2 * x * 2y³ + ( 2y³)²] = x² + 4xy³ + 4y^6 >>>>resposta

Answer 2

$\large\boxed{\begin{array}{l}\tt a)~\sf(3+x)^2=3^2+2\cdot3\cdot x+x^2\\\sf(3+x)^2=9+6x+x^2\\\tt b)~\sf(x+5)^2=x^2+2\cdot x\cdot5+5^2\\\sf(x+5)^2=x^2+10x+25\\\tt c)~\sf(x+y)=x^2+2xy+y^2\\\tt d)~\sf(x+2)^2=x^2+2\cdot x\cdot2+2^2\\\sf(x+2)^2=x^2+4x+4\\\tt e)~\sf(3x+2)^2=(3x)^2+2\cdot3x\cdot2+2^2\\\sf(3x+2)^2=9x^2+12x+4\\\tt f)~\sf(2x+1)^2=(2x)^2+2\cdot2x\cdot1+1^2\\\sf(2x+1)^2=4x^2+4x+1\\\tt g)~\sf(a+1)^2=a^2+2\cdot a\cdot 1+1^2\\\sf (a+1)^2=a^2+2a+1\\\tt h)~\sf(y^3+3)^2=(y^3)^2+2\cdot y^3\cdot3+3^2\\\sf(y^3+3)^2=y^6+6y^3+9\\\tt i)~\sf(a+b)^2=a^2+2ab+b^2\\\tt j)~\sf(x+2y^3)^2=x^2+2\cdot x\cdot2y^3+(2y^3)^2\\\sf(x+2y^3)^2=x^2+4xy^3+4y^6\end{array}}$