Log3(5х – 6) - log72= 3; log0,5 (2х + 1) = -2; log2 (4-2x) + log23= 1; log7(x-l)=log72 + log73; 1 ≤7х-3< 49; log2 (1 - 2х) < 0; lg (0,5x - 4) < 2; log0,2 (2х+3) ≥ -3;
236
464
Ответы на вопрос:
2) log0,5_(2x+1) = - 2; - log2_(2x+1) = - 2; log2_(2x+1) = 2; 2x+ 1= 2^2; 2x = 3; x= 1,5. 3)log2_(4 - 2x) + log2_3 = 1; -2x)*3 = 1; log2_(12 - 6x) = 1; 12 - 6x = 2^1; 12 - 6x = 2; - 6x = -10; x = 10/6= 5/3. 4) log7_(x-1) = log7_2 + log7_3; log7_(x-1) = log7_(2*3); x - 1 = 6; x = 7. 5)1 ≤ 7x - 3 < 49; +3 1 + 3 ≤ 7x < 49 + 3; 4 ≤ 7x < 52; 4/7 ≤ x < 52/7. 6) log2_(1 - 2x) < 0; log2_(1 - 2x) < log2_1; 2 > 1; ⇒ 1 - 2x < 1; - 2x < 1 - 1; - 2x < 0; /-2 < 0; x > 0 7) lg(0,5 x - 4) < 2; lg(0,5x - 4) < lg100; 0,5x - 4 < 100; 0,5 x < 104; * 2> 0; x < 208 8) log0,2_(2x+3) ≥ - 3; 0,2 = 1/5 = 5^(-1); - log5_(2x + 3) ≥ - 3; /-1 < 0; log5_(2x + 3) ≤ 3; log5_(2x+3) ≤ log5_125; 5 > 1; ⇒ 2x + 3 ≤ 125; 2 x ≤ 122; x ≤ 61. в первом не понятно условие.
F(x) ' =(x^2+4x-6/x+3 )' = (x^2) '+(4x)' - (6/x)' +(3 )' = 2x +4 +6/x^2 +0 = 2x +4 +6/x^2 найдем значение производной в заданной точке x0=-2f '(-2)= 2*(-2) +4 +6/(-2)^2 = 3/2 ответ f '(-2) = 3/2
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