Задача 34748 Задание на картинке...

milanazhavor

21 марта 2019 г. в 00:46

Задание на картинке

sova

21 марта 2019 г. в 09:36

1.
Основное логарифмическое тождество:
a^{log _ab}=b
a>0;a ≠ 1; b>0

(1–2²)=1–4=–3
Логарифмы отрицательных чисел не существуют
Применить формулу нельзя.

2.
(8^2/5)^1/2=8^(2/5)·(1/2)=8^1/5
∛64=∛4³=4
8^1/5·4=(2³)^1/5·2²=2^3/5·2²=2^(3/5)+2=2^13/5

3.
√14=√7·√2

√14·(√7+2–3√2)=

=√7·√7·√2+2√14–3√7·√2·√2=

=7·√2 +2sqrt·(14)–6√7

4.
a⁶+b⁶=(a²)³+(b²)³=
формула (m³+n³=(m+n)·(m²–mn+n²);
m=a²; n=b²)

=(a²+b²)·((a²)²–a²·b²+(b²)²)=(a²+b²)·(a⁴–a²b²+b⁴)

(a⁶+b⁶)/(a⁴–a²b²+b⁴)=a²+b²

5a)
(1+tg²30 °)·(1–tg²30 °)–ctg²60 °=

=(1+(√3/3)²)·(1–(√3/3)²)·–(√3/3)²=

=(1+(1/3))·(1–(1/3)) – (1/3)=(1²–(1/3)²– (1/3)=(8/9)–(6/9)=2/9

5b)
(sin(π/6) – cos(π/6))²+tg(7π/4)=

= sin²(π/6)–2sin(π/6)·cos(π/6)+ cos²(π/6)+tg(2π–(π/4))=

= 1– sin(2·(π/6))– tg(π/4)=1–sin(π/3)+1=–sin(π/3)= –√3/2;

6a)
1–cos² α =sin² α

1–cos² α +sin² α =sin² α +sin² α =2sin² α

ctg(5π/4)=ctg(π+(π/4))=ctg(π/4)=1

(1–cos² α +sin² α)·сtg(5 π/4)=2sin² α ·1=2sin² α

6b)
2sin² α+cos² α =sin² α +sin² α +cos² α =

=sin² α +(sin² α +cos² α)=sin² α +1

sin⁴ α –1=(sin² α –1)·(sin² α +1)

(2sin² α+cos² α)·(sin² α –1)/(tg α·(sin⁴ α –1))=

=(sin² α +1)·(sin² α –1)/(tg α·(sin² α –1)·(sin² α +1))=

=1/tg α =ctg α

При α =π/4

ctg(π/4)=1

6c

sin(– α ) =– sin α

sin(π– α )=sin α

tg((π/2)+ α )= – ctgα

sin(– α )+sin(π– α )·(tg((π/2)+ α )=

=–sin α +sin α ·(cos α /sin α )=–sin α +cos α

7.
cos1230 °= cos( 4·360 ° – 210 °)=

=cos(–210 °)= cos(–180 ° – 30 °)=

=–cos(30 °)=–√3/2:

sin(–405 °)=–sin405 ° =

=– sin( 360 ° + 45 °)=–sin 45 °= – √2/2:

tg(–7π/3)=–tg(7π/3)=–tg(2π+(π/3))= – tg(π/3)= – √3;

ctg(29π/6)=ctg((30π/6)–(π/6))=ctg(5π– (π/6))=ctg(–π/6)=– √3.