1) (13-9)(13+9)= 4*22=88
2)(20-19)(20+19)=1*39=39
3) (11/5)2-(14/5)2= 121/25 - 196/25= -3
4)(7/2)2 - (37/10)2= 49/4 - 1369/100= - 36/25 = -1 11/25= -1,44
5)25/36 - 4/9=1/4=0,25
6) 49/81-1/36=187/324
7) 25/144-9/16=-7/18
8) 9/100-16/25= -11/20=-0,55
9)64/225-16/25=-16/45
10) 0
11) (10/3)2-(21/5)2=100/9-441/25= - 1469/225 = - 6 119/225
12) (31/6)2-(22/3)2= 951/36-484/9= - 325/12= - 27 1/12
13) (51-41)(51+41)=10*92=920
14) (54-46)(54+46)= 8*100=800
15)(76-24)(76+24)=52*100=5200
16)(328-172)(328+172)=156*500=78000
17)(11/3)2-(7/3)2=121/9-49/9=8
18) (68/9)2-(40/9)2= 4624/81-1600/81=112/3=37,3
Объяснение:
Объяснение:
1) Решениеy=(4·x-9)^5
((4·x-9)^5)' = 20(4·x-9^)4
Поскольку:
((4·x-9)5)' = 5·(4·x-9)^5-^1((4·x-9))' = 20(4·x-9)^4
(4·x-9)' = 4
20(4·x-9)^4
y=(x2-3x+1)7
2) Решение:((x2-3x+1)7)' = (-7·3x·ln(3)+14·x)(x2-3x+1)6
Поскольку:
((x2-3x+1)7)' = 7·(x2-3x+1)7-1((x2-3x+1))' = (-7·3x·ln(3)+14·x)(x2-3x+1)6
(x2-3x+1)' = (x2)' + (-3x)' + (1)' = 2·x + (-3x·ln(3)) = -3x·ln(3)+2·x
(x2)' = 2·x2-1(x)' = 2·x
(x)' = 1
Здесь:
Решение ищем по формуле:
(af(x))' = af(x)*ln(a)*f(x)'
(-3x)' = -3x·ln(3)(x)' = -3x·ln(3)
(x)' = 1
(-7·3x·ln(3)+14·x)(x2-3x+1)6
3) Решение:y=(sin(x))^3
(sin(x)^3)' = 3·sin(x)^2·cos(x)
Поскольку:
(sin(x)^3)' = 3·(sin(x))^3-1((sin(x)))' = 3·sin(x)^2·cos(x)
(sin(x))' = cos(x)
3·sin(x)2·cos(x)