Question

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20 (ESCCAI) A solução do sistema (x + y = 9 : log, x x + log, y = 3 a) (4,5). (5.4) d) (0.9). (9,0) b) (1.8), (8.1) c) (10.-1).(-1, 10) c) (3,6). (6,3)​

Answer 1

Resposta:

b

Explicação passo a passo:

Vamos pegar a segunda equação

$log_2x+log_2y=3$

$log_2x=3-log_2y$

Usando a propriedade dos logaritmos temos:

$x=2^{3-log_2y}$

Simplificando:

$x=2^3*2^{-log_2y}$

$x=8*y^{-1}$

$x=\frac{8}{y}$

Inserindo o valor de x na primeira equação:

$\frac{8}{y}+y=9$

$\frac{8+y^2=9y}{y}$

$y^2-9y+8=0$

Podemos fatorar essa expressão da seguinte forma:

$(y-1)(y-8)$

Logo as raízes são (1,8)e(8,1)

Answer 2

$\large\boxed{\begin{array}{l}\sf (\bf {ESCCAI})\sf A\,soluc_{\!\!,}\tilde ao\,do\,sistema\\\begin{cases}\sf x+y=9\\\sf \ell og_2x+\ell og_2y=3\end{cases}~\sf\acute e:\\\sf a) (4,5),(5,4 )~~d)(0,9),( 9,0)\\\sf b)(1,8),(8,1)~~e)(10,-1),(-1,10)\\\sf c)(3,6),(6,3)\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\begin{cases}\rm x+y=9\\\rm\ell og_2x+\ell og_2y=3\end{cases}\\\rm \ell og_2x+\ell og_2y=3\\\rm \ell og_2(x\cdot y)=3\longrightarrow x\cdot y=2^3\\\rm x\cdot y=8\\\begin{cases}\rm x+y=9\\\rm x\cdot y=8\end{cases}\\\\\begin{cases}\rm y=9-x\\\rm x\cdot y=8\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\rm x\cdot (9-x)=8\\\rm 9x-x^2=8\\\rm x^2-9x+8=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-9)^2-4\cdot1\cdot8\\\rm\Delta=81-32\\\rm\Delta=49\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{-(-9)\pm\sqrt{49}}{2}\\\\\rm x=\dfrac{9\pm7}{2}\begin{cases}\rm x_1=\dfrac{9+7}{2}=\dfrac{16}{2}=8\\\\\rm x_2=\dfrac{9-7}{2}=\dfrac{2}{2}=1\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\rm se\,x=8:\\\rm y=9-x\\\rm y=9-8\\\rm y=1\\\rm se\,x=1:\\\rm y=9-x\\\rm y=9-1\\\rm y=8\\\rm S=\{(1,8),(8,1)\}\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~b}}}}}\end{array}}$