Resolva o sistema de equações: Obs: respostas sem sentido ou brincadeiras serão eliminadas.... Pergunta de ideia deKrikor

December 2019 1 141 Report

Resolva o sistema de equações:

$\left\{\begin{matrix} \mathtt{2^x\cdot 4^y=\dfrac{3}{4}}\quad \qquad\\ \\ \mathtt{y^3-\dfrac{x\cdot y^2}{2}=0} \end{matrix}\right.$

Obs: respostas sem sentido ou brincadeiras serão eliminadas

Lista de comentários

Lukyo

Verified answer

Resolver o sistema de equações.

$\left\{\!\begin{array}{lc}\mathsf{2^x\cdot 4^y=\dfrac{3}{4}}&\quad\mathsf{(i)}\\\\ \mathsf{y^3-\dfrac{x\cdot y^2}{2}=0}&\quad\mathsf{(ii)}\end{array}\right.$

Da equação (i), tiramos

$\mathsf{2^x\cdot 4^y=\dfrac{3}{4}}\\\\\\ \mathsf{2^x\cdot (2^2)^y=\dfrac{3}{4}}\\\\\\ \mathsf{2^x\cdot 2^{2y}=\dfrac{3}{4}}\\\\\\ \mathsf{2^{x+2y}=\dfrac{3}{4}}$

Logo,

$\mathsf{x+2y=\ell og_2\left(\dfrac{3}{4}\right)}\\\\\\ \mathsf{x=\ell og_2\left(\dfrac{3}{4}\right)-2y}\\\\\\ \mathsf{x=a-2y\qquad\quad (iii)}$

sendo $\mathsf{a=\ell og_2\,\dfrac{3}{4}.}$

Rearrumando a equação (ii), temos

$\mathsf{y^3-\dfrac{xy^2}{2}=0}\\\\\\ \mathsf{2y^3-xy^2=0}\\\\ \mathsf{y^2 \cdot (2y-x)=0}$

Substituindo (iii) acima,

$\mathsf{y^2 \cdot \big[2y-(a-2y)\big]=0}\\\\ \mathsf{y^2 \cdot (2y-a+2y)=0}\\\\ \mathsf{y^2 \cdot (4y-a)=0}\\\\ \mathsf{y^2=0~~~ou~~~4y-a=0}\\\\ \mathsf{y=0~~~ou~~~y=\dfrac{a}{4}}$

=====

• Para y = 0, encontramos

$\mathsf{x=a-2y}\\\\ \mathsf{x=a-2\cdot 0}\\\\ \mathsf{x=a} \\\\ \mathsf{x=\ell og_2\,\dfrac{3}{4}}$

• Para $\mathsf{y=\dfrac{a}{4}=\dfrac{1}{4}\,\ell og_2\,\dfrac{3}{4},}$ encontramos

$\mathsf{x=a-2y}\\\\ \mathsf{x=a-2\cdot \dfrac{a}{4}}\\\\\\ \mathsf{x=a-\dfrac{a}{2}}\\\\\\ \mathsf{x=\dfrac{a}{2}}\\\\\\ \mathsf{x=\dfrac{1}{2}\,\ell og_2\,\dfrac{3}{4}}$

Conjunto solução: $\mathsf{S=\left\{\left(\ell og_2\,\dfrac{3}{4},\,0\right),\, \left(\dfrac{1}{2}\,\ell og_2\,\dfrac{3}{4},\, \dfrac{1}{4}\,\ell og_2\,\dfrac{3}{4} \right) \right\}}$

Bons estudos! :-)

Krikor Excelente! Muito obrigado, Lukyo! :)

Lukyo Por nada :)

Lista de comentários

Verified answer

Resolva essa conta:

Recomendar perguntas

Helpful Links

Helpful Social