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⠀⠀⠀☞ X = ± 2 · √5. ✅
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⠀⠀⠀➡️⠀Inicialmente vamos relembrar que raízes podem ser reescritas como o denominador do expoente da base, de forma a simplificar os cálculos. Na equação do enunciado temos portanto:
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[tex]\LARGE\blue{\text{$\sf X^{^2} = \sqrt[2]{10^{^2}} + \sqrt[2]{10^{^2}}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X^{^2} = 10^{^{\frac{2}{2}}} + 10^{^{\frac{2}{2}}}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X^{^2} = 10^{^1}+ 10^{^1}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X^{^2} = 10+ 10 = 20$}}[/tex]
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⠀⠀⠀➡️⠀Aplicando a radiciação em ambos os lados da igualdade temos:
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[tex]\LARGE\blue{\text{$\sf \sqrt[2]{\sf X^{^2}} = \pm \sqrt[2]{20}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X^{^{\frac{2}{2}}} = \pm \sqrt[2]{4 \cdot 5}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X^{^1} = \pm \sqrt[2]{4} \cdot \sqrt[2]{5}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X = \pm \sqrt[2]{2^{^2}} \cdot \sqrt{5}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X = \pm 2^{^{\frac{2}{2}}} \cdot \sqrt{5}$}}[/tex]
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[tex]\LARGE\blue{\text{$\sf X = \pm 2^{^1} \cdot \sqrt{5}$}}[/tex]
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[tex]\huge\green{\boxed{\rm~~~\gray{X}~\pink{=}~\blue{\pm 2 \cdot \sqrt{5} }~~~}}[/tex] ✅
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[tex]\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]☁
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⠀⠀⠀☀️ L͎̙͖͉̥̳͖̭̟͊̀̏͒͑̓͊͗̋̈́ͅeia mais sobre radiciação:
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https://brainly.com.br/tarefa/38363792 ✈
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[tex]\huge\blue{\text{\bf\quad Bons~estudos.}}[/tex] ☕
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[tex]\quad\qquad(\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios})[/tex] ☄
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[tex]\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }\LaTeX}[/tex]✍
[tex]\sf(\purple{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly}[/tex] ☘☀❄☃☂☻)
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[tex]\Huge\green{\text{$\underline{\red{\mathbb{S}}\blue{\mathfrak{oli}}~}~\underline{\red{\mathbb{D}}\blue{\mathfrak{eo}}~}~\underline{\red{\mathbb{G}}\blue{\mathfrak{loria}}~}$}}[/tex] ✞
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