ответ:
d=b^2-4ac=(-1)^2-4*1*(-72)=1+288=\sqrt{289}
289
=17
х1=\frac{-b- \sqrt{d} }{2a} = \frac{1-17}{2} = \frac{-16}{2} =-8
2a
−b−
d
=
2
1−17
=
2
−16
=−8
х2=\frac{-b+ \sqrt{d} }{2a} = \frac{1+17}{2} = \frac{18}{2} = 9
2a
−b+
d
=
2
1+17
=
2
18
=9
ответ: -8 и 9
d=b^2-4ac=7^2-4*(-4)*(-3)=49-48=\sqrt{1} =1
1
=1
х1=\frac{-b- \sqrt{d} }{2a} = \frac{-7-1}{2*(-4)} = \frac{-8}{-8} =1
2a
−b−
d
=
2∗(−4)
−7−1
=
−8
−8
=1
х2=\frac{-b+ \sqrt{d} }{2a} = \frac{-7+1}{(-8)} = \frac{-6}{-8} =0,75
2a
−b+
d
=
(−8)
−7+1
=
−8
−6
=0,75
1.а) * * * cosα*cosβ = ( cos(α -β) +cos(α+β) ) /2
* * * sinα*cosβ = ( sin(α+β) +sin(α - β) ) /2
cos18°*cos72° - sin63°cos27° =
( cos(72°- 18°) +cos(72°+18°) ) /2 - ( sin(63°+27°) +sin(63° -27°) ) /2 =
(cos54° + cos90° )/2 - ( sin90° +sin36 ) /2 =
( cos(90°- 36°) + 0 )/2 - ( 1 +sin36°)/2 = ( sin36° - (1+sin36°) ) /2 = - 1/2.
- - - - - - -
2. вычислить cos10°*cos50°*cos70°
можно разными комбинациями
cos10°*cos50°*cos70° =(1/2) ( cos(50°+10°) +cos(50°- 10°) )*cos70°=
(1/2) (cos60° +cos40°)*cos70° =(1/2) ( (1/2)*cos70° ° +cos40°*cos70° )=
=(1/2) ( (1/2)*cos70° +(1/2)( cos(70°+40°)+cos(70°-40°) ) ) =
(1/4) ( cos70° +cos110° + cos30° ) =1/4) ( cos70° +cos(180° -70°) + √3 /2 )
(1/4) ( cos70°- cos70° + √3 /2 ) = √3 /8
- - - - - - -
3. упростите
cos(α+β)*cos(α -β) - cos²α -cos²β =
(1/2) ( cos(α+β+α -β ) +cos(α+β- α +β) ) + - cos²α - cos²β =
(1/2)cos2α+ (1/2)cos2β - cos²α - cos²β =
(1/2)( cos²α - sin²α + cos²β - sin²β - 2cos²α - 2cos²β ) =
(1/2)( - cos²α - sin²α - sin²β - cos²β ) =
(-1/2) ( (cos²α + sin²α) + (sin²β+ cos²β) ) = (-1/2) ( 1 + 1 ) = - 1
Второй
cos(α+β)*cos(α -β) - cos²α -cos²β =
cos(α+β)*cos(α -β) - (1+cos2α) /2 - (1+cos2β) /2 =
cos(α+β)*cos(α -β) - 1 - (cos2α + cos2β) /2 =
cos(α+β)*cos(α -β) - 1 - (2cos(α + β) *cos(α- β) ) /2 =
cos(α+β)*cos(α -β) - 1 - cos(α + β) *cos(α- β) = - 1
