Question

determine o valor numérico das seguintes expressões algébricas:​

Answer 1

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$\LARGE\green{\boxed{\rm~~~\red{A)}~\gray{f(5)}~\pink{=}~\blue{ 48 }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{B)}~\gray{f(2, -5)}~\pink{=}~\blue{ -5 }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{C)}~\gray{f(7)}~\pink{=}~\blue{ 11,2 }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{D)}~\gray{f(-3,-4)}~\pink{=}~\blue{ 3 }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{E)}~\gray{f(-3)}~\pink{=}~\blue{ 49 }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{F)}~\gray{f\left(\dfrac{-1}{2}, \dfrac{2}{3}\right)}~\pink{=}~\blue{ 3 }~~~}}$

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

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☺lá, Gabriel, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo, feita através de algumas manipulações algébricas. ✌

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

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$\LARGE\gray{\boxed{\sf\blue{ (a + 1) \cdot (a + 3) }}}$

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➡ $\large\blue{\sf f(5) = (5 + 1) \cdot (5 + 3) }$

$\large\blue{\sf = 6 \cdot 8 }$

$\large\blue{\sf = 48 }$

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$\huge\green{\boxed{\rm~~~\gray{f(5)}~\pink{=}~\blue{ 48 }~~~}}$ ✅

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

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$\LARGE\gray{\boxed{\sf\blue{ 4x + 2z -3 }}}$

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➡ $\large\blue{\sf f(2, -5) = 4 \cdot 2 + 2 \cdot (-5) - 3 }$

$\large\blue{\sf = 8 - 10 - 3 }$

$\large\blue{\sf = -5 }$

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$\huge\green{\boxed{\rm~~~\gray{f(2, -5)}~\pink{=}~\blue{ -5 }~~~}}$ ✅

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

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$\LARGE\gray{\boxed{\sf\blue{ \dfrac{3x^2 - 5x}{x + 3} }}}$

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➡ $\large\blue{\sf f(7) = \dfrac{3 \cdot 7^2 - 5 \cdot 7}{7 + 3} }$

$\large\blue{\sf =  \dfrac{3 \cdot 49 - 35}{10} }$

$\large\blue{\sf =  \dfrac{147 - 35}{10} }$

$\large\blue{\sf =  \dfrac{112}{10} }$

$\large\blue{\sf =  11,2 }$

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$\huge\green{\boxed{\rm~~~\gray{f(7)}~\pink{=}~\blue{ 11,2 }~~~}}$ ✅

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

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$\LARGE\gray{\boxed{\sf\blue{ \dfrac{x^2 - 3y}{y^2 + 3x} }}}$

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➡ $\large\blue{\sf f(-3,-4) = \dfrac{(-3)^2 - 3 \cdot (-4)}{(-4)^2 + 3 \cdot (-3)} }$

$\large\blue{\sf = \dfrac{9 + 12}{16 - 9} }$

$\large\blue{\sf = \dfrac{21}{7} }$

$\large\blue{\sf = 3 }$

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$\huge\green{\boxed{\rm~~~\gray{f(-3,-4)}~\pink{=}~\blue{ 3 }~~~}}$ ✅

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

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$\LARGE\gray{\boxed{\sf\blue{ 4x^2 - 5x - 2 }}}$

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➡ $\large\blue{\sf f(-3) = 4 \cdot (-3)^2 - 5 \cdot (-3) - 2 }$

$\large\blue{\sf = 4 \cdot 9 + 15 - 2 }$

$\large\blue{\sf = 36 + 15 - 2 }$

$\large\blue{\sf = 49 }$

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$\huge\green{\boxed{\rm~~~\gray{f(-3)}~\pink{=}~\blue{ 49 }~~~}}$ ✅

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

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$\LARGE\gray{\boxed{\sf\blue{ 6a + 9b }}}$

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➡ $\large\blue{\sf f\left(\dfrac{-1}{2}, \dfrac{2}{3}\right) = 6 \cdot \dfrac{-1}{2} + 9 \cdot \dfrac{2}{3}}$

$\large\blue{\sf = \dfrac{6 \cdot (-1)}{2} + \dfrac{9 \cdot 2}{3} }$

$\large\blue{\sf = \dfrac{-6}{2} + \dfrac{18}{3} }$

$\large\blue{\sf = \dfrac{-6}{2} \cdot 1 + \dfrac{18}{3} \cdot 1}$

$\large\blue{\sf = \dfrac{-6}{2} \cdot \dfrac{3}{3} + \dfrac{18}{3} \cdot \dfrac{2}{2}}$

$\large\blue{\sf = \dfrac{-6 \cdot 3}{2 \cdot 3} + \dfrac{18 \cdot 2}{3 \cdot 2}}$

$\large\blue{\sf = \dfrac{-18}{6} + \dfrac{36}{6} }$

$\large\blue{\sf = \dfrac{-18 + 36}{6} }$

$\large\blue{\sf = \dfrac{18}{6} }$

$\large\blue{\sf = 3 }$

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$\huge\green{\boxed{\rm~~~\gray{f\left(\dfrac{-1}{2}, \dfrac{2}{3}\right)}~\pink{=}~\blue{ 3 }~~~}}$ ✅

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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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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$