Question

ПОМОГИТЕ ПОЖАЛУЙСТА!
срочно

Answer 1

1

$2sin^2a+cos2a=sin^2a+sin^2a+cos^2a-sin^2a=1$

2

$lim_{x->3}\frac{x^2-5x+6}{3x^2-9x}=lim_{x->3}\frac{(x-2)(x-3)}{3x(x-3)} = lim_{x->3}\frac{x-2}{3x}=\frac{1}{9}$

3

$f(x)=\sqrt{4x-1}$

$f'(x)=\frac{1}{2\sqrt{4x-1}}*4=\frac{2}{\sqrt{4x-1}}$

Answer 2

1) =sin^2a+sin^2a +1-2sin^2a= 1

2)x^2-5x+6=0

D=25-24=1

x1=(5+1)/2= 6/2=3

x2=(5-1)/2=4/2=2

lim(x->3) (x-3)(x-2)/3x(x-3) = lim(x->3) (x-2)/3x = 3-2/3*3= 1/9

3)f'(x) = 1/2√(4x-1) * (4x-1) ' =4/2√(4x-1)