Question

Agora vamos resolver o exercício;
Calcule:
a) (2x + 3)²=

b) (4a² + 5b³) =

c) (y - 1)²=

d) (6 - 4y³) =

e) (6x + 7)(6x -7) =

f) (11 - 8y⁵)(11+ 8y⁵)=

Answer 1

Resposta:

Explicação passo-a-passo:

Para resolver essas questoes devera ConheCer os seguintes Casos notaveis

$(x+y)^{2}=x^{2} +2xy+y^{2} \\(x-y)^{2}=x^{2} -2xy+y^{2} \\x^{2} +y^{2} =(x+y)(x-y)$

a) (2x + 3)²=

4x²+12x+9

b) (4a² + 5b³)² =$16x^{4} +40a^{2}b^{3} +25b^{6}$

c) (y - 1)²=y²-2y+1

d) (6 - 4y³)² =36-48

y³+16$y^{6}$

e) (6x + 7)(6x -7) =

36x²-49

f) (11 - 8y⁵)(11+ 8y⁵)=121-64$y^{25}$

Answer 2

Quadrado da soma de dois termos

$\large\boxed{\boxed{\boxed{\boxed{\mathsf{(a+b)^2=a^2+2ab+b^2}}}}}$

Quadrado da diferença de dois termos

$\large\boxed{\boxed{\boxed{\boxed{\mathsf{(a-b)^2=a^2-2ab+b^2}}}}}$

Produto da soma pela diferença de dois termos

$\large\boxed{\boxed{\boxed{\boxed{\mathsf{(a-b)\cdot(a+b)}}}}}$

Cubo da soma de dois termos

$\mathsf{(a+b)^3=a^3+3a^2b+3ab^2+b^3}$

Cubo da diferença de dois termos

$\mathsf{(a-b)^3=a^3-3a^2b+3ab^2-b^3}$

$\dotfill$

a)

$\mathsf{(2x +3)^2=(2x)^2+2\cdot2x\cdot3+(3)^2}\\\mathsf{(2x+3)^2=4x^2+12x+9}$

b)

$\mathsf{ (4a^2 + 5b)^3=(4a^2)^3+3\cdot(4a^2)^2\cdot5b+3\cdot4a^2\cdot(5b)^2+(5b)^3}\\\mathsf{(4a^2+5b)^3=64a^6+3\cdot16a^4\cdot5b+3\cdot4a^2\cdot25b^2+125b^3}\\\mathsf{(4a^2+5b)^3=64a^6+240a^4b+300a^2b^2+125b^3}$

c)

$\mathsf{(y - 1)^2=(y)^2-2\cdot y\cdot1+1^2}\\\mathsf{(y-1)^2=y^2-2y+1}$

d)

$\mathsf{(6 - 4y)^3=6^3-3\cdot6^2\cdot4y+3\cdot6\cdot(4y)^2-(4y) ^3}\\\mathsf{(6-4y)^3=216-3\cdot36\cdot4y+3\cdot6\cdot16y^2-64y^3}\\\mathsf{(6-4y)^3=216-432y+288y^2-64y^3}$

e)

$\mathsf{(6x + 7)(6x -7)=(6x)^2-7^2}\\\mathsf{(6x+7)(6x-7)=36x^2-49}$

f)

$\mathsf{(11 - 8y^5)(11+ 8y^5)=11^2-(8y^5)^2}\\\mathsf{121-64y^{10}}$

$\dotfill$

$\displaystyle\mathsf{\ell ife=\int\limits_{birth}^{death}\dfrac{happiness}{time}dtime}$