Question

Combinari de x-2 luate cate 2=21

Answer 1

${C ^{2} _{x-2}} = 21$

$x \: \in \: \mathbb{N}$

$x - 2 > 2 = > x > 4$

$\frac{(x - 2)!}{2!(x - 2 - 2)!} = 21$

$\frac{(x - 2)!}{2(x - 4)!} = 21$

$\frac{(x - 2)(x - 3)(x - 4)!}{2(x - 4)!} = 21$

$\frac{(x - 2)(x - 3)}{2} = 21$

$(x - 2)(x - 3) = 2 \times 21$

$(x - 2)(x - 3) = 42$

${x}^{2} - 3x - 2x + 6 = 42$

${x}^{2} - 3x - 2x + 6 - 42=0$

${x}^{2} - 5x - 36 = 0$

${x}^{2} - 9x + 4x - 36 = 0$

$x(x - 9) + 4(x - 9) = 0$

$(x - 9)(x + 4) = 0$

$x - 9 = 0 = > x = 9 > 0$

$x + 4 = 0 = > x = - 4 < 0 \: nu \: convine$

$= > x = 9$

Answer 2

C.E.x≥2
(x-2) (x-3)/2=21
(x-2)(x-3)=(x-3)(x-2)=42=6*7
x-3=6
x-2=7
x=9∈Domeniului de existenta
(x-2)(x-3) crescatoare pt x>0, x=9 solutie unica
as simple as that!!