Прошу ! сравните с нулем: 1) f'(2), если, [tex]f(x) = \frac{ {3}^{1 - 2x} }{ {x}^{ - 4} } [/tex]2) [tex]f( \frac{1}{3} )[/tex] если, f(x) = ln(2 - 3x) +x

1) cosα · tgα - sinα = cosα · sinα/cosα - sinα = sinα - sinα = 0

2) cos2α - ? ctg2α - ? sinα = -2/3, 3π/2 < α < 2π - это iv четверть, где sinα < 0, cosα > 0.

cos2α = cos²α - sin²α = 1 - 2sin²α = 1 - 2 · (-2/3)² = 1 - 2 · 4/9 = 1 - 8/9 = 1/9

sin2α = 2sinα · cosα = 2 · (-2/3) · √5/3 = -4√5/9

cos²α = 1 - sin²α = 1 - (-2/3)² = 1 - 4/9 = 5/9; cosα = √(5/9) = √5/3

ctg2α = cos2α/sin2α = 1/9 : (-4√5/9) = -1/(4√5)

3) tg195° = tg(180° + 15°) = tg15° = tg(45° - 30°) =

= (tg45° - tg30°)/(1 + tg45° · tg30°) = (1 - √3/3)/(1 + 1 · √3/3) =

= (1 - √3/3)/(1 + √3/3) = (1 - √3/3)²/((1 + √3/3)(1 - √3/3)) =

= (1 + 3/9 - 2√3/3)/( 1 -3/9) = (1 + 1/3 - 2√3/3)/(1 - 1/3) = (4/3 -2√3/3)/(2/3) =

= (4 - 2√3)/3 : (2/3) = 2(2 - √3)/3 · 3/2 = 2 - √3

c1=-1

c26=49

s26=(a1+a26)/2*26=13(48)=624

х+19/ х-5 = -1

обе части умножим на (х-5)

х+19=-х+5

2х=5-19

х=-7