Question

Alguém pode me ajudar por favor?
Urgente,valendo 100 pontoooos!

Answer 1

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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☺lá, Chay, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌

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☔ Para cada um dos exercícios abaixo o que faremos basicamente será substituir o valor os valores dos pares/ ternas ordenadas nas equações e verificar se a igualdade será mantida/ encontrar o valor desconhecido/ verificar se (n-1) termos substituídos resultam no n-ésimo termo.

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1)Ⓐ___________________________✍

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➡ $\sf\blue{ 4 \cdot (1) - (2) + (-5) = 0 ?}$

❌ $\sf\blue{ -3 \neq 0 }$  ❌

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Ⓑ_____________________________✍

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➡ $\sf\blue{ -(1) + (2) - 3 \cdot (-5) = 0 ? }$

❌ $\sf\blue{16 \neq 0}$  ❌

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Ⓒ_____________________________✍

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➡ $\sf\blue{ 2 \cdot (1) - 3 \cdot (2) - (-5) = 0 ? }$

❌ $\sf\blue{ 1 \neq 0 }$  ❌

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Ⓓ_____________________________✍

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➡ $\sf\blue{ \dfrac{(1)}{4} + (2) - \dfrac{(-5)}{2} = 8 ?}$

$\sf\blue{ = \dfrac{1}{4} + \dfrac{8}{4} + \dfrac{10}{4} }$

❌ $\sf\blue{ \dfrac{19}{4} \neq 8 }$  ❌

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$\large\green{\boxed{\rm~~~\red{ 1)}~\blue{ N.D.O. }~~~}}$ ✅

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2)_____________________________✍

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$\large\gray{\boxed{\rm\blue{  }}}$

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➡ $\sf\blue{ 5 \cdot (3) + (-4) + (5) = 4 ? }$

❌ $\sf\blue{ 16 \neq 4 }$  ❌

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$\large\green{\boxed{\rm~~~\red{ 2)}~\blue{ N\tilde{a}o~\acute{e}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$ ✅

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3)Ⓐ___________________________✍

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➡ $\sf\blue{ (1) - 3 \cdot (-2p) - (1) = 4}$

➡ $\sf\blue{ 6p = 4 }$

➡ $\sf\blue{ p = \dfrac{4}{6} }$

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$\large\green{\boxed{\rm~~~\red{ 3)~a)}~\gray{p}~\pink{=}~\blue{\dfrac{2}{3}}~~~}}$ ✅

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Ⓑ_____________________________✍

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 ➡ $\sf\blue{ (p + 3) - 3 \cdot (3) - (0) = 4}$

➡ $\sf\blue{ p + 3 = 13 }$

➡ $\sf\blue{ p = 10 }$

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$\large\green{\boxed{\rm~~~\red{ 3)~b)}~\gray{p}~\pink{=}~\blue{10}~~~}}$ ✅

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Ⓒ_____________________________✍

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 ➡ $\sf\blue{ (p) - 3 \cdot (p) - (p + 5) = 4}$

➡ $\sf\blue{ p - 3p - p = 9 }$

➡ $\sf\blue{ p = \dfrac{-9}{3} }$

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$\large\green{\boxed{\rm~~~\red{ 3)~c)}~\gray{p}~\pink{=}~\blue{-3}~~~}}$ ✅

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4)_____________________________✍

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➡ $\sf\blue{ k \cdot (-1) - (0) - 3 \cdot (1) = -5 }$

➡ $\sf\blue{ -k  - 3 = -5 }$

➡ $\sf\blue{ k = 2 }$

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$\large\green{\boxed{\rm~~~\red{ 4)}~\gray{k}~\pink{=}~\blue{ 2 }~~~}}$ ✅

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5)Ⓐ___________________________✍

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☔ O maior monômio tem grau 1: A

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$\large\green{\boxed{\rm~~~\red{ 5)~a)}~\blue{ A }~~~}}$ ✅

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Ⓑ_____________________________✍

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☔ O maior monômio tem grau 1: A

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$\large\green{\boxed{\rm~~~\red{ 5)~b)}~\blue{ A }~~~}}$ ✅

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Ⓒ_____________________________✍

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☔ O maior monômio tem grau 1: A

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$\large\green{\boxed{\rm~~~\red{ 5)~c)}~\blue{ A }~~~}}$ ✅

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Ⓓ_____________________________✍

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☔ O maior monômio tem grau 2: B

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$\large\green{\boxed{\rm~~~\red{ 5)~d)}~\blue{ B }~~~}}$ ✅

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Ⓔ_____________________________✍

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 ☔ O maior monômio tem grau 2: B

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$\large\green{\boxed{\rm~~~\red{ 5)~e)}~\blue{ B }~~~}}$ ✅

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Ⓕ_____________________________✍

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 ☔ O maior monômio tem grau 2: B

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$\large\green{\boxed{\rm~~~\red{ 5)~f)}~\blue{ B }~~~}}$ ✅

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6)_____________________________✍

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➡ $\sf\blue{ 3 \cdot (3) - 2 \cdot (k) = 5}$

➡ $\sf\blue{ 9 - 2k = 5}$

➡ $\sf\blue{ -2k = -4}$

➡ $\sf\blue{ k = \dfrac{4}{2}}$

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$\large\green{\boxed{\rm~~~\red{ 6)}~\gray{k}~\pink{=}~\blue{ 2 }~~~}}$ ✅

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7)_____________________________✍

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➡ $\sf\blue{ -2 \cdot (2a) + 3 \cdot (a - 1) + 2 \cdot (a) = 0 }$

➡ $\sf\blue{ -4a + 3a - 3 + 2a = 0 }$

➡ $\sf\blue{ a = 3 }$

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$\large\green{\boxed{\rm~~~\red{ 7)}~\gray{a}~\pink{=}~\blue{ 3 }~~~}}$ ✅

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8)Ⓐ___________________________✍

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➡ $\sf\blue{ 2 \cdot (1) + (2) + 5 \cdot (3) = 15 ? }$

❌ $\sf\blue{ 19 \neq 15 }$ ❌

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$\large\green{\boxed{\rm~~~\red{ 8)}~\blue{ N\tilde{a}o~\acute{e}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$ ✅

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Ⓑ_____________________________✍

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 ➡ $\sf\blue{ 2 \cdot (0) + 7 \cdot (0) - 3 \cdot (0) = 0 ? }$

✅  $\sf\blue{ 0 = 0 }$ ✅

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.$\large\green{\boxed{\rm~~~\red{ 8)}~\blue{ \acute{E}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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