Question

Ajudeeemmm pooor favooooorrrr

Answer 1

Explicação passo-a-passo:

3)

$\left|\begin{array}{ccc}x&x\\5&x\\\end{array}\right|=-6\\\\x^{2} -5x=-6\\\\x^{2} -5x+6=0\\\\\Delta=(-5)^{2}-4.1.6=25-24=1\\\\x'=\dfrac{-(-5)+\sqrt{1} }{2.1}\\\\x'=\dfrac{5+1}{2}=\dfrac{6}{2} \\\\\boxed{x'=3}\\\\x''=\dfrac{-(-5)-\sqrt{1} }{2.1}\\\\x''=\dfrac{5-1}{2}=\dfrac{4}{2} \\\\\boxed{x''=2}$

$\boxed{\boxed{S\{2,3\}}}$

4)

$\left|\begin{array}{ccc}1&0&1\\2&1&x\\0&1&2\end{array}\right| =0\\\\\text Usando\ a\ Regra\ de\ Sarrus:\\\\\left|\begin{array}{ccc}1&0&1\\2&1&x\\0&1&2\end{array}\right|\left\begin{array}{ccc}1&0\\2&1\\0&1\end{array}\right| =0\\\\1.1.2+0.x.0+1.2.1-0.2.2-1.x.1-1.1.0=0\\\\2+0+2-0-x-0=0\\\\4-x=0\\\\\boxed{\boxed{x=4}}$

Resposta D

5)

$\left|\begin{array}{ccc}2&1&3\\4&-1&5\\4&2&x\end{array}\right| =0\\\\\text Usando\ a\ Regra\ de\ Sarrus:\\\\\left|\begin{array}{ccc}2&1&3\\4&-1&5\\4&2&x\end{array}\right|\left\begin{array}{ccc}2&1\\4&-1\\4&2\end{array}\right|=0\\\\2.(-1).x+1.5.4+3.4.2-1.4.x-2.5.2-3.(-1).4=0\\\\-2x+20+24-4x-20+12=0\\\\-6x+36=0\\\\-6x=-36\\\\x=\dfrac{-36}{-6}\\\\\boxed{\boxed{x=6}}$

Resposta D