Question

faz com as contas a 6 e a 7 ​

Answer 1

Resposta:

Olá!

Explicação passo a passo:

$(a+b)^2=a^2+2.a.b+b^2$

6) a)

$(3+x)^2=3^2+2.3.x+x^2\\(3+x)^2=x^2+6x+9$

b)

$(5+3x)^2=5^2+2.5.3x+(3x)^2\\(5+3x)^2=9x^2+30x+25$

c)

$(2x+y)^2=(2x)^2+2.2x.y+y^2\\(2x+y)^2=4x^2+4xy+y^2$

d)

$(r+4s)^2=r^2+2.r.4s+(4s)^2\\(r+4s)^2=r^2+8rs+16s^2$

e)

$(10x+y)^2=(10x)^2+2.10x.y+y^2\\(10x+y)^2=100x^2+20xy+y^2$

$(a-b)^2=a^2-2.a.b+b^2$

7.a)

$(3x-2y)^2=(3x)^2-2.3x.2y+(2y)^2\\(3x-2y)^2=9x^2-12xy+4y^2$

b)

$(2x-b)^2=(2x)^2-2.2x.b+b^2\\(2x-b)^2=4x^2-4xb+b^2$

c)

$(5x^2-1)^2=(5x^2)^2-2.5x^2.1+1^2\\(5x^2-1)^2=25x^4-10x^2+1$

d)

$(x^2-1)^2=(x^2)^2-2.x^2.1+1^2\\(x^2-1)^2=x^4-2x^2+1$

e)

$(9x^2-1)^2=(9x^2)^2-2.9x^2.1+1^2\\(9x^2-1)=81x^4-18x^2+1$

Bons Estudos!