Question

$2^{x^{2}-6x+0.5}=frac{1}{16sqrt{2}}$

Answer 1

$2^{x^2-6x-0.5}=frac{1}{2^{frac{9}{2}}$

$2^{x^2-6x-0.5}=2^{frac{9}{2}$

$x^2-6x-0.5={frac{9}{2}$

$2x^2-12x-0,5+9=0$

$2x^2-12x+10=0$

D=144-2*4*10=64

x1=5

x2=1

Ответ: 5,1

Answer 2

2^(x^2-6x+0,5)=1/(16*sqrt(2))

2^(x^2-6x+0,5)=2^(-9,5)

x^2-6x+0,5=-4,5

x^2-6x+5=0

D=b^2-4ac=16

x1,2=(-b±sqrt(D))/2

x1=(6-4)/2=1

x2=(6+4)/2=5