Question

Resolva aplicando o Produto de Stevin

A) (y-11) . (y+6)

B)

$\sf \left(x-\frac{2}{5} \right)\cdot\left(x-\frac{3}{2} \right)$
com explicação passo-a-passo

Answer 1

$\LARGE\text{ \underline{\sf Ol\acute{a}{,}\ boa\ noite!} }$

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☃️ $\large\text{ \underline{\sf Produtos\ not\acute{a}veis.} }$

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Para se calcular um produto notável, existem diversas fórmulas, irei lhe mostrar todas elas. ... Vamos lá! :)

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Todas as Fórmulas:

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Quadrado da soma de dois termos:

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$\large\begin{array}{ll}\sf F\acute{o}rmula:\\\\ \sf (A+ B)^2 = A^2 +2\cdot A\cdot B+ B^2\\\\ \sf Comprovando\ a\ f\acute{o}rmula:\\\\ \sf (A+B)^2= (A+B) \cdot (A+B)\\\\ \sf Aplicando\ a\ distributiva:\\\\\sf A^2 + AB +AB+B^2\\\\ \sf Comprovamos\ a\ f\acute{o}rmula:\\\\ \sf A^2 + 2\cdot A \cdot B + B^2\end{array}$

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Quadrado da diferença de dois termos:

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$\large\begin{array}{ll}\sf F\acute{o}rmula:\\\\ \sf (A- B)^2 = A^2 -2\cdot A\cdot B+ B^2\\\\ \sf Comprovando\ a\ f\acute{o}rmula:\\\\ \sf (A- B)^2= (A- B) \cdot (A- B)\\\\ \sf Aplicando\ a\ distributiva:\\\\\sf A^2 - AB -AB+B^2\\\\ \sf Comprovamos\ a\ f\acute{o}rmula:\\\\ \sf A^2 - 2\cdot A \cdot B + B^2\end{array}$

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Produto da soma pela diferença de dois termos:

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$\large\begin{array}{ll}\sf F\acute{o}rmula:\\\\ \sf (A+B)\cdot (A-B)= A^2 -B^2\\\\ \sf Comprovando\ a\ f\acute{o}rmula:\\\\ \sf (A+B)\cdot (A-B)=\\\\ \sf Aplicando\ a\ distributiva:\\\\\sf A^2 \not{-AB} \not{+AB-B^2}\\\\ \sf Comprovamos\ a\ f\acute{o}rmula:\\\\ \sf A^2 -B^2\end{array}$

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Cubo da soma de dois termos:

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$\large\begin{array}{ll}\sf F\acute{o}rmula:\\\\ \sf (A+B)^3=A^3+ 3\cdot A^2 \cdot B + 3\cdot A\cdot B^2 + B^3\\\\ \sf Comprovando\ a\ f\acute{o}rmula:\\\\ \sf (A+B)^2 \cdot (A+B)=\\\\ \sf Resolvendo\ (A+B)^2:\\\\\sf ( A^2 + 2\cdot A \cdot B + B^2)=\\\\ \sf Fazendo\ a\ distributiva:\\\\ \sf ( A^2 + 2\cdot A \cdot B + B^2) \cdot (A+B)=\\ \sf A^3+ 3\cdot A^2 \cdot B + 3\cdot A\cdot B^2 + B^3\\\\ \sf Comprovamos\ a\ f\acute{o}rmula: \\\\ \sf A^3+ 3\cdot A^2 \cdot B + 3\cdot A\cdot B^2 + B^3\end{array}$

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Cubo da diferença de dois termos:

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$\large\begin{array}{ll}\sf F\acute{o}rmula:\\\\ \sf (A-B)^3=A^3- 3\cdot A^2 \cdot B + 3\cdot A\cdot B^2 - B^3\\\\ \sf Comprovando\ a\ f\acute{o}rmula:\\\\ \sf (A-B)^2 \cdot (A-B)=\\\\ \sf Resolvendo\ (A-B)^2:\\\\\sf ( A^2 - 2\cdot A \cdot B +B^2)=\\\\ \sf Fazendo\ a\ distributiva:\\\\ \sf ( A^2 - 2\cdot A \cdot B + B^2) \cdot (A-B)=\\ \sf A^3- 3\cdot A^2 \cdot B + 3\cdot A\cdot B^2 - B^3\\\\ \sf Comprovamos\ a\ f\acute{o}rmula: \\\\ \sf A^3- 3\cdot A^2 \cdot B + 3\cdot A\cdot B^2 - B^3\end{array}$

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Quadrado da soma de 3 números:

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$\large\begin{array}{ll}\sf F\acute{o}rmula:\\\\ \sf (A+B+C)^2= A^2 +B^2+C^2+2AB+2AC+ 2BC \\\\ \sf Comprovando\ a\ f\acute{o}rmula:\\\\ \sf (A+B+C) \cdot (A+B+C)=\\\\ \sf Aplicando\ a\ distributiva:\\\\\sf (A+B+C)\cdot (A+B+C)=\\\sf A^2 +B^2+C^2+2AB+2AC+ 2BC \\\\ \sf Comprovamos\ a\ f\acute{o}rmula:\\\\ \sf A^2 +B^2+C^2+2AB+2AC+ 2BC\end{array}$

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$\large\boxed{\boxed{\sf Vamos\ agora\ para\ sua\ quest\tilde{a}o:}}$

Alternativa A)

$\large\boxed{\sf A)\ (y-11) . (y+6)}$

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 Resolvendo:

$\large\boxed{\boxed{\boxed{\begin{array}{ll}\sf (y-11) . (y+6)\\\\ \sf Aplicando\ a\ distributiva:\\\\\sf yy+y\cdot \:6+\left(-11\right)y+\left(-11\right)\cdot \:6\\ \sf yy+6y-11y-11\cdot \:6\\\\\boxed{ \sf =y^2-5y-66}\end{array}}}}$

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Alternativa B)

$\large\boxed{\sf B)\left(x- \frac{2}{5}\right) \cdot \left(x- \frac{3}{2}\right) }$

    $\searrow$

 Resolvendo:

$\large\boxed{\boxed{\boxed{\begin{array}{ll}\sf \left(x- \frac{2}{5}\right) \cdot \left(x- \frac{3}{2}\right)\\\\ \sf Aplicando\ a\ distributiva:\\\\\sf xx+x\left(-\frac{3}{2}\right)+\left(-\frac{2}{5}\right)x+\left(-\frac{2}{5}\right)\left(-\frac{3}{2}\right)\\\\ \sf xx-\frac{3}{2}x-\frac{2}{5}x+\frac{2}{5}\cdot \frac{3}{2} \\\\ \sf \boxed{\sf =x^2-\frac{19x}{10}+\frac{3}{5}} \end{array}}}}$

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⭐ Espero ter ajudado!

⭐ Bons estudos!

$\LARGE\begin{matrix} \underbrace{ \sf By: Pedro } \end{matrix}$