Question

Ajudem pooor favooooorrrr

Answer 1

Explicação passo-a-passo:

6)

$\left|\begin{array}{ccc}8&x+2\\-1&x\\\end{array}\right|=\left|\begin{array}{ccc}x+2&-8\\x&x-2\\\end{array}\right|\\\\\\8.x-(x+2).(-1)=(x+2).(x-2)-(-8).x\\\\8x+x+2=x^{2} -4+8x\\\\x+2=x^{2} -4\\\\x^{2} -x-4-2=0\\\\x^{2} -x-6=0$

Nessa equação de segundo grau a soma $S=\frac{-b}{a}=\frac{-(-1)}{1}=1$ e o produto

$P=\frac{c}{a} =\frac{-6}{1}=-6$

$x'=3$ e $x''=-2$

Resposta B

7)

$\left|\begin{array}{ccc}x+1&3&2\\3&x&1\\x&2&4\end{array}\right|=-29\\\\\\\left|\begin{array}{ccc}x+1&3&2\\3&x&1\\x&2&4\end{array}\right|\left\begin{array}{ccc}x+1&3\\3&x\\x&2\end{array}\right| =-29\\\\\\(x+1).x.4+3.1.x+2.3.2-3.3.4-(x+1).1.2-2.x.x=-29\\\\4x^{2} +4x+3x+12-36-2x-2-2x^{2} =-29\\\\2x^{2} +5x-26=-29\\\\2x^{2} +5x-26+29=0\\\\2x^{2} +5x+3=0$

Δ=5²-4.2.3  =25-24=1

$x'=\frac{-5+\sqrt{1} }{2.2} \\\\x'=\frac{-5+1}{4}\\\\x'=\frac{-4}{4} \\\\x'=-1$

$x''=\frac{-5-\sqrt{1} }{2.2}\\\\x''=\frac{-5-1}{4}\\\\x''=\frac{-6}{4}\\\\x''=\frac{-3}{2}$