31,75; 508
Объяснение:
(an) - арифметическая прогрессия
a₁+a₂+a₃=27
a₁+a₁+d+a₁+2d=27
3(a₁+d)=27
a₁+d=9
a_1+d=a₂ => a₂=9
a₁+9+a₃=27
a₁+a₃=27-9=18
a₃=18-a₁
(bn) - геометрическая прогрессия
b₁=a₁-1
b₂=a₂-1=9-1=8
b₃=a₃+3=18-a₁+3=21-a₁
8/(a₁-1) = (21-a₁)/8
(a₁-1)(21-a₁)=64
21a₁-21-a₁²+a₁-64=0
-a₁²+22a₁-85=0
a₁²-22a₁+85=0
D=(-22)²-4*1*85= 484-340=144=12²
(a₁)₁ = (22+12)/2 = 34/2 = 17
(a₁)₂ = (22-12)/2 = 10/2 = 5
Получаем сразу две геометрические прогрессии:
1) b₁=17-1=16, b₂=8, b₃=21-17=4 => q = 8/16=1/2
S₇ = b₁(q⁷-1)/(q-1) = 16((1/2)⁷-1)/(1/2 -1) = 16(1/128 -1)/(-1/2) =
= -16*2*(-127/128)=127/4 = 31,75
2) b₁=5-1=4, b₂=8, b₃=21-5=16 => q=8/4=2
S₇ = b₁(q⁷-1)/(q-1) = 4(2⁷-1)/(2-1) = 4*(128-1)/1 = 4*127 = 508
5^(x-2) = 5^0 2^(x² -3x +8) = 2^6
x-2 = 0 x² -3x +8 = 6
x = 2 x² -3x +2 = 0
2) 3·4^x =48 x = 1 и х = 2
4^x = 16 6)7^(2x-8)·7^(x+7) = 0
4^x = 4² нет решений
x=2 7)(0,2)^x ≤ 25·5√5
3)3^x=27·3√9 5^-x ≤ 5²·5·5^1/2
3^x = 3³·3·3 5^-x ≤5^3,5
3^x = 3^5 -x ≤ 3,5
x = 5 x ≥ -3,5
4)3^x + 3^(x +1) = 4 8)(1/2)^-x + 2^(3 +x) ≤9
3^x(1 +3) = 4 2^x +2^(3 +x) ≤ 9
3^x·4 = 4 2^x(1 +2^3) ≤ 9 | :9
3^x = 1 2^x ≤ 1
x = 0 2^x ≤2^0
x≤ 0
31,75; 508
Объяснение:
(an) - арифметическая прогрессия
a₁+a₂+a₃=27
a₁+a₁+d+a₁+2d=27
3(a₁+d)=27
a₁+d=9
a_1+d=a₂ => a₂=9
a₁+9+a₃=27
a₁+a₃=27-9=18
a₃=18-a₁
(bn) - геометрическая прогрессия
b₁=a₁-1
b₂=a₂-1=9-1=8
b₃=a₃+3=18-a₁+3=21-a₁
8/(a₁-1) = (21-a₁)/8
(a₁-1)(21-a₁)=64
21a₁-21-a₁²+a₁-64=0
-a₁²+22a₁-85=0
a₁²-22a₁+85=0
D=(-22)²-4*1*85= 484-340=144=12²
(a₁)₁ = (22+12)/2 = 34/2 = 17
(a₁)₂ = (22-12)/2 = 10/2 = 5
Получаем сразу две геометрические прогрессии:
1) b₁=17-1=16, b₂=8, b₃=21-17=4 => q = 8/16=1/2
S₇ = b₁(q⁷-1)/(q-1) = 16((1/2)⁷-1)/(1/2 -1) = 16(1/128 -1)/(-1/2) =
= -16*2*(-127/128)=127/4 = 31,75
2) b₁=5-1=4, b₂=8, b₃=21-5=16 => q=8/4=2
S₇ = b₁(q⁷-1)/(q-1) = 4(2⁷-1)/(2-1) = 4*(128-1)/1 = 4*127 = 508