log(2) (4^x + 4) = x + log(2) (2^x*2^1 - 3)
log(2) (4^x + 4) = x + log(2) (2^(x+1) - 3)
ОДЗ
4^x + 4 > 0 x∈ R
2^(x+1) > 3
log(2) 2^(x+1) > log(2) 3
x + 1 > log(2) 3
x > log(2) 3 - 1 ≈ 1.59 - 1 ≈ 0.59
ОДЗ x ∈ (log(2) 3 - 1 , +∞ )
log(2) (4^x + 4) = x + log(2) (2^(x+1) - 3)
log(2) (4^x + 4) = log (2) 2^x + log(2) (2^(x+1) - 3)
log(2) (4^x + 4) = log(2) 2^x*(2*2^x - 3)
снимаем логарифмы
4^x + 4 = 2^x*(2*2^x - 3)
(2^x)^2 + 4 = 2*2^x*2^x - 3*2^x
(2^x)^2 - 3*2^x - 4 = 0
2^x = t > 0
t^2 - 3t - 4 = 0
D=9 + 16 = 25 = 5²
t₁₂ = (3 +- 5)/2 = -1 4
1. t₁ = -1
решений нет t>0
2. t=4
2^x = 4
x = 2 (входит в ОДЗ x > log(2) 3 - 1 )
ответ х=2
Решение
4/x ≥ 0
x ≤ 4
-----------------////////////////////////////////-------------->
- ∞ 4 x
x∈( - ∞ ; 4]
1) 4 1/10 - 3 4/15 = 3 33/30 - 3 8/30 = 25/30 = 5/6
2) 5/6 * 5/6 = 25/36
3) 4/10:1 1/5 = 4/10 * 5/6 = 2/6 = 1/3
4) 25/36 + 1/3 = 25/36 + 12/36 = 37/36 = 1 1/36
(1/4)^-2= 16, дробь переворачивается, а показатель меняет знак
<span>y=f(x) где f(x)=x</span>²<span>.
f(x</span>²<span>)=(х</span>²)²=х⁴<span>
f(x</span>²<span>-2)=(х</span>²-2)²=х⁴-4х²+4
<span>f(x</span>³<span>)=(х</span>³)²=х⁶
<span>f(x</span>³<span>+x)=(х</span>³+х)²=х⁶+2х⁴+х²
Вроде так.