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krickayavika96
@krickayavika96
July 2022
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nafanya2014
Verified answer
Sin(π-α)=sinα
cos(2π-α)=cosα
cos(α-π)=cos(π-α)=-cosα
sin(α-(π/2))=-sin(π/2-α)=-cosα
ответ. А) -1
х²-2х-1=t
3x²-6x-3=3t
t²+3t-10=0
D=49
t=2 или t=-5
x²-2x-1=2 или х²-2х-1=-5
х²-2х-3=0 х²-2х+4=0
D=16 D<0 уравнение не имеет корней
x=-1; x=3
Ответ. A) -1; 3
1 votes
Thanks 1
oganesbagoyan
Verified answer
14)
sin(π -α)*ctq(-α)*cos(2π -α)/ (cos(α -π)*sin(α -π/2) ) =
sinα*(-ctqα)*cosα)/ cos(-(π-α))*sin (-(π/2-
α) ) =
-sinα*ctqα*cosα)/ cos(π-α)*(-sin (π/-
α) ) =
(-cos²α)/ (-cosα*(-cosα) ) = -1. →A)
-------
15)
(x² -2x -1)² +(3x² -6x-13) =0 ;
(x² -2x -1)² +3(x² -2x-1) -10 =0 ;
t² +3t -10 =0 ;
D =3² -4*1(-10) =49 =7.
t₁ = (-3 -7)/2 = -5 ⇒x² -2x -1= -5⇔
x² -2x +4=0(не имеет вещ. корней).
t₂ = (-3 +7)/2 = 2 ⇒x² -2x -1=2 ⇔x² -2x -3 =0⇒ [x= -1 ; x=3.
→A)
0 votes
Thanks 1
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Answers & Comments
Verified answer
Sin(π-α)=sinαcos(2π-α)=cosα
cos(α-π)=cos(π-α)=-cosα
sin(α-(π/2))=-sin(π/2-α)=-cosα
ответ. А) -1
х²-2х-1=t
3x²-6x-3=3t
t²+3t-10=0
D=49
t=2 или t=-5
x²-2x-1=2 или х²-2х-1=-5
х²-2х-3=0 х²-2х+4=0
D=16 D<0 уравнение не имеет корней
x=-1; x=3
Ответ. A) -1; 3
Verified answer
14)sin(π -α)*ctq(-α)*cos(2π -α)/ (cos(α -π)*sin(α -π/2) ) =
sinα*(-ctqα)*cosα)/ cos(-(π-α))*sin (-(π/2-α) ) =
-sinα*ctqα*cosα)/ cos(π-α)*(-sin (π/-α) ) =
(-cos²α)/ (-cosα*(-cosα) ) = -1. →A)
-------
15)
(x² -2x -1)² +(3x² -6x-13) =0 ;
(x² -2x -1)² +3(x² -2x-1) -10 =0 ;
t² +3t -10 =0 ;
D =3² -4*1(-10) =49 =7.
t₁ = (-3 -7)/2 = -5 ⇒x² -2x -1= -5⇔x² -2x +4=0(не имеет вещ. корней).
t₂ = (-3 +7)/2 = 2 ⇒x² -2x -1=2 ⇔x² -2x -3 =0⇒ [x= -1 ; x=3. →A)