Question

Resolva a equação do arranjo

Answer 1

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$\tt b)~\sf A_{x,2}+A_{x-1,2}=162\\\dfrac{x!}{(x-2)!}+\dfrac{(x-1)!}{(x-1-2)!}=162\\\sf\dfrac{x\cdot(x-1)\cdot\diagup\!\!\!\!(x-\diagup\!\!\!\!2)!}{\diagup\!\!\!\!(x-\diagup\!\!\!\!2)!}+\dfrac{(x-1)\cdot(x-2)\cdot\diagup\!\!\!\!(x-\diagup\!\!\!\!3)!}{\diagup\!\!\!\!(x-\diagup\!\!\!\!\!3)!}=162\\\sf x\cdot(x-1)+(x-1)\cdot(x-2)=162\\\sf x^2-x+x^2-2x-x+2-162=0\\\sf 2x^2-4x-160=0\div 2\\\sf x^2-2x-80=0$

$\sf \Delta=b^2=4ac\\\sf\Delta=(-2)^2-4\cdot1\cdot(-80)\\\sf\Delta=4+320\\\sf\Delta=324\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-2)\pm\sqrt{324}}{2\cdot1}\\\sf x=\dfrac{2\pm18}{2}\begin{cases}\sf x_1=\dfrac{2+18}{2}=\dfrac{20}{2}=10\\\sf x_2=\dfrac{2-18}{2}=-\dfrac{16}{2}=-8\end{cases}\\\sf como~estamos~falando~de~fatorial,o~resultado~ser\acute a~positivo\\\sf portanto\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=10\checkmark}}}}$

Answer 2

b) Ax,2 + Ax-1,2 = 162

x!/(x-2)! + (x-1)!/(x-1-2)! = 162

x × (x-1) × (x-2)!/(x-2)! + (x-1) × (x-2) × (x-3)!/(x-3)! = 162

x × (x-1) + (x-1) × (x-2) = 162

(x-1) × (x+x-2) = 162

(x-1) × (2x-2) = 162

(x-1) × 2(x-1) = 162

(x-1)² × 2 = 162

2(x-1)² = 162

2(x²-2x+1) = 162

x² - 2x + 1 = 81

x² - 2x + 1 - 81 = 0

x² - 2x - 80 = 0

∆ = (-2)² - 4.1.(-80)

∆ = 4 + 320

∆ = 324

$x = \frac{ - ( - 2) + - \sqrt{324} }{2 \times 1}$

$x = \frac{2 + - 18}{2}$

$x = \frac{2 + 18}{2} = 10$

$x = \frac{2 - 18}{2} = - 8$

Precisamos do resultado positivo, então 10.