Question

genteeeeee me ajudaa ​

Answer 1

✅ De acordo com o estudo das técnicas de fatoração, segue abaixo os resultados

a) ½

b) 3

c) (y + 3)/(x + 7)

d) (x + 1)/(x - 1)

e) 5/3(x - 2)

f) (3a² + b)/(a + 2)

 

✍️ Solução: Basta aplicar algumas fatorações conhecidas, como: produtos notáveis e diferenças de quadrados. Essas manipulações vem diretamente das propriedades operatórias dos números reais, tais como as leis associativas, distributivas e comutativas.

a)

$\large\begin{array}{lr}\begin{aligned}\rm \dfrac{(x+2)^2}{2x^2+8x+8} &=\rm \dfrac{(x+2)^2}{2(x^2+4x+4)}, ~x\neq-2\\\\&=\rm \dfrac{ \cancel{(x+2)^2}}{2 \cancel{(x+2)^2}} \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \dfrac{(x+2)^2}{2x^2+8x+8} = \dfrac{1}{2} }}}}\end{array}$

 

b)

$\large\begin{array}{lr}\begin{aligned}\rm \dfrac{3x^2+24x+48}{(x+4)^2} &=\rm \dfrac{3(x^2+8x+16)}{(x+4)^2} ,~x\neq-4\\\\&=\rm \dfrac{ \cancel{3(x+4)^2}}{2 \cancel{(x+4)^2}} \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \dfrac{(x+2)^2}{2x^2+8x+8} = 3 }}}}\end{array}$

 

c)

$\large\begin{array}{lr}\begin{aligned}\rm \dfrac{3x+21+xy+7y}{(x+7)^2} &=\rm \dfrac{3(x+7)+y(x+7)}{(x+7)^2} ,~x\neq-7\\\\&=\rm \dfrac{\bcancel{(x+7)}(y+3)}{(x+4)^{\bcancel{2}}} \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \dfrac{3x+21+xy+7y}{(x+7)^2} = \dfrac{y+3}{x+7} }}}}\end{array}$

 

d)

$\large\begin{array}{lr}\begin{aligned}\rm \dfrac{x^2-1}{x^2-2x+1} &=\rm \dfrac{(x+1) \cancel{(x-1)}}{(x-1)^{\cancel{2}}} ,~x\neq1 \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \dfrac{x^2-1}{x^2-2x+1} = \dfrac{x+1}{x-1} }}}}\end{array}$

 

e)

$\large\begin{array}{lr}\begin{aligned}\rm \dfrac{15x-30}{9x^2-36x+36} &=\rm \dfrac{15(x-2)}{9(x^2-4x+4)^2} ,~x\neq2\\\\&=\rm \dfrac{15\bcancel{(x-2)}}{9(x-2)^{\bcancel{2}}} \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \dfrac{15x-30}{9x^2-36x+36} = \dfrac{5}{3(x-2)} }}}}\end{array}$

 

f)

$\large\begin{array}{lr}\begin{aligned}\rm \dfrac{(3a^2+b)^2}{6a^2+3a^3+2b+ab} &=\rm \dfrac{(3a^2+b)^2}{3a^2(a+2)+b(a+2)} ,~a>0,~b>0\\\\&=\rm \dfrac{(3a^2+b)^{\cancel{2}}}{(a+2)\cancel{(3a^2+b)}} \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \dfrac{(3a^2+b)^2}{6a^2+3a^3+2b+ab} = \dfrac{3a^2+b}{a+2} }}}}\end{array}$

 

✔️ Resolvido.

 

⚓️️️️ Seção de links para complementar o estudo sobre álgebra:

• brainly.com.br/tarefa/771780

$\rule{7cm}{0.01mm}\\\texttt{Bons estudos! :D}\\\rule{7cm}{0.01mm}$