a) [tex3]1[/tex3]
b) [tex3]2[/tex3]
c) [tex3]4[/tex3]
d) [tex3]8[/tex3]
e) [tex3](2/3)a[/tex3]
Resposta
Gabarito : [tex3]A[/tex3]
Ensino Fundamental ⇒ Fatoração Tópico resolvido
-
menelaus
- Mensagens: 628
- Registrado em: 14 Mar 2013, 10:13
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- Agradeceram: 16 vezes
Jun 2014
05
01:36
Fatoração
Se [tex3]a[/tex3]
é um número real maior que [tex3]1/8[/tex3]
, o valor de [tex3]\sqrt[3]{a+\frac{a+1}{3}\sqrt {\frac{8a-1}{3}}}[/tex3]
+ [tex3]\sqrt[3]{a-\frac{a+1}{3}\sqrt {\frac{8a-1}{3}}}[/tex3]
é igual a :
a) [tex3]1[/tex3]
b) [tex3]2[/tex3]
c) [tex3]4[/tex3]
d) [tex3]8[/tex3]
e) [tex3](2/3)a[/tex3]
Gabarito : [tex3]A[/tex3]
a) [tex3]1[/tex3]
b) [tex3]2[/tex3]
c) [tex3]4[/tex3]
d) [tex3]8[/tex3]
e) [tex3](2/3)a[/tex3]
Gabarito : [tex3]A[/tex3]
Editado pela última vez por caju em 24 Out 2017, 01:20, em um total de 2 vezes.
Razão: TeX --> TeX3
Razão: TeX --> TeX3
-
VALDECIRTOZZI
- Mensagens: 2569
- Registrado em: 04 Ago 2008, 17:08
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- Agradeceram: 1590 vezes
Jun 2014
09
12:23
Re: Fatoração
Temos que:
[tex3]x=\sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}[/tex3]
[tex3]x^3=\left(\sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^3[/tex3]
[tex3]x^3=\left(\sqrt[3]{a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}\right)^3+3\cdot \left(\sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^2\cdot \sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}+3\cdot \sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\cdot\left(\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^2+\left(\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^3[/tex3]
[tex3]x^3=a+\cancel{\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}}+a\cancel{-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}+3 \cdot \sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2\cdot \left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)}+3\cdot\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot \left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)}+\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot \left[a^2-\left(\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2\right]}[/tex3]
[tex3]+ \sqrt[3]{\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot \left[a^2-\left(\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2\right]}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left[a^2-\left(\frac{a^2+2a+1}{9}\right)\cdot \left(\frac{8a-1}{3}\right)\right]}[/tex3]
[tex3]+\sqrt[3]{\left(a-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left[a^2-\left(\frac{a^2+2a+1}{9}\right)\cdot \left(\frac{8a-1}{3}\right)\right]}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left(\frac{27a^2-8a^3+a^2-16a^2+2a-8a+1}{27}\right)}[/tex3]
[tex3]+\sqrt[3]{\left(a-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left(\frac{27a^2-8a^3+a^2-16a^2+2a-8a+1}{27}\right)}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{-8a^3+12a^2-6a+1}{27}\right)}[/tex3]
[tex3]+\sqrt[3]{\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{-8a^3+12a^2-6a+1}{27}\right)}[/tex3]
[tex3]\frac{x^3-2a}{3}=-\sqrt[3]{\left(a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{2a-1}{27}\right)^3}-\sqrt[3]{\left(a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{2a-1}{27}\right)^3}[/tex3]
[tex3]-\frac{x^3-2a}{3}=\left(\frac{2a-1}{3}\right)\cdot\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\left(\frac{2a-1}{3}\right)\cdot\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/tex3]
[tex3]-\frac{x^3-2a}{3}=\left(\frac{2a-1}{3}\right)\underbrace{ \left(\sqrt[3]{\left( a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)}+ \sqrt[3]{ a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)}[/tex3]
[tex3]-x^3+2a=\left(2a-1\right)\cdot x[/tex3]
[tex3]x^3+2ax-x-2a=0[/tex3]
[tex3]\left(x-1\right)\left(x^2+x+2a\right)=0[/tex3]
[tex3]x-1=0 \Longleftrightarrow x=1[/tex3]
Espero ter ajudado!
[tex3]x=\sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}[/tex3]
[tex3]x^3=\left(\sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^3[/tex3]
[tex3]x^3=\left(\sqrt[3]{a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}\right)^3+3\cdot \left(\sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^2\cdot \sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}+3\cdot \sqrt[3]{a+\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\cdot\left(\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^2+\left(\sqrt[3]{a-\frac{a+1}{3} \cdot\sqrt{\frac{8a-1}{3}}}\right)^3[/tex3]
[tex3]x^3=a+\cancel{\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}}+a\cancel{-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}+3 \cdot \sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2\cdot \left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)}+3\cdot\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot \left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)}+\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot \left[a^2-\left(\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2\right]}[/tex3]
[tex3]+ \sqrt[3]{\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot \left[a^2-\left(\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)^2\right]}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left[a^2-\left(\frac{a^2+2a+1}{9}\right)\cdot \left(\frac{8a-1}{3}\right)\right]}[/tex3]
[tex3]+\sqrt[3]{\left(a-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left[a^2-\left(\frac{a^2+2a+1}{9}\right)\cdot \left(\frac{8a-1}{3}\right)\right]}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left(\frac{27a^2-8a^3+a^2-16a^2+2a-8a+1}{27}\right)}[/tex3]
[tex3]+\sqrt[3]{\left(a-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}\right)\left(\frac{27a^2-8a^3+a^2-16a^2+2a-8a+1}{27}\right)}[/tex3]
[tex3]\frac{x^3-2a}{3}=\sqrt[3]{\left(a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{-8a^3+12a^2-6a+1}{27}\right)}[/tex3]
[tex3]+\sqrt[3]{\left(a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{-8a^3+12a^2-6a+1}{27}\right)}[/tex3]
[tex3]\frac{x^3-2a}{3}=-\sqrt[3]{\left(a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{2a-1}{27}\right)^3}-\sqrt[3]{\left(a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)\cdot\left(\frac{2a-1}{27}\right)^3}[/tex3]
[tex3]-\frac{x^3-2a}{3}=\left(\frac{2a-1}{3}\right)\cdot\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\left(\frac{2a-1}{3}\right)\cdot\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/tex3]
[tex3]-\frac{x^3-2a}{3}=\left(\frac{2a-1}{3}\right)\underbrace{ \left(\sqrt[3]{\left( a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)}+ \sqrt[3]{ a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)}[/tex3]
[tex3]-x^3+2a=\left(2a-1\right)\cdot x[/tex3]
[tex3]x^3+2ax-x-2a=0[/tex3]
[tex3]\left(x-1\right)\left(x^2+x+2a\right)=0[/tex3]
[tex3]x-1=0 \Longleftrightarrow x=1[/tex3]
Espero ter ajudado!
Editado pela última vez por caju em 24 Out 2017, 01:25, em um total de 2 vezes.
Razão: TeX --> TeX3
Razão: TeX --> TeX3
So many problems, so little time!
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