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1) 2* 25^x - 5^(x+1) * 2^x + 2^(2x+1) =<0
2)1+2log_x+2_5 = log_5(x+2)
1) 2* 25^x - 5^(x+1) * 2^x + 2^(2x+1) =<0
2)1+2log_x+2_5 = log_5(x+2)
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1) 2*5^(2x) - 5*5^x*2^x + 2*2^(2x) <= 0Делим все на 2^(2x)
2*(5/2)^(2x) - 5*(5/2)^x + 2 <= 0
Замена (5/2)^x = y
2y^2 - 5y + 2 <= 0
(y - 2)(2y - 1) <= 0
y = (5/2)^x ∈ [1/2; 2]
x ∈ [log_(5/2) (1/2); log_(5/2) (2)]
2) 1 + 2log_(x+2) (5) = log_5 (x+2)
По свойствам логарифмов log_a(b) = 1/log_b(a), поэтому замена
y = log_5(x+2); log_(x+2) (5) = 1/y
1 + 2/y = y
y^2 - y - 2 = 0
(y + 1)(y - 2) = 0
y1 = log_5(x+2) = -1; x1 = -2 + 5^(-1) = -2 + 1/5 = -1,8
y2 = log_5(x+2) = 2; x2 = -2 + 5^2 = -2 + 25 = 23