Решите логарифмическое уравнение : lg(x^2+1)-lg(x-2)=1 Решите показательное уравнение :...

1)
$lg( {x}^{2} + 1) - lg(x - 2) = 1 \\ lg( {x}^{2} + 1) = 1 + lg(x - 2) \\ lg( {x}^{2} + 1) = lg10 + lg(x - 2) \\ lg( {x}^{2} + 1) = lg(10(x - 2)) \\ {x}^{2} + 1 = 10(x - 2) \\ {x}^{2} + 1 = 10x - 20 \\ {x}^{2} - 10 + 21 = 0 \\ d = {b}^{2} - 4ac = 100 - 4 \times 21 = 100 - 84 = 16 \\ x1 = \frac{10 + 4}{2} = \frac{14}{2} = 7 \\ x2 = \frac{10 - 4}{2} = \frac{6}{2} = 3$
ОДЗ:
x^2 + 1 > 0, x^2 > -1 => x € R
x - 2 > 0, x > 2
Оба корня удовл.ОДЗ., значит, являются корнями уравнения.

Ответ: 3; 7.

2)
0 \\ t + \frac{125}{t} = 30 \\ {t}^{2} - 30t + 125 = 0 \\ d = {b}^{2} - 4ac = 900 - 4 \times 125 = 400 \\ t1 = \frac{30 + 20}{2} = \frac{50}{2} = 25 \\ t2 = \frac{30 - 20}{2} = \frac{10}{2} = 5 \\ \\ {5}^{x} = 25 \\ x = 2 \\ \\ {5}^{x} = 5 \\ x = 1" alt=" {5}^{x} + \frac{125}{ {5}^{x} } = 30 \\ {5}^{x} = t. \: \: \: \: t > 0 \\ t + \frac{125}{t} = 30 \\ {t}^{2} - 30t + 125 = 0 \\ d = {b}^{2} - 4ac = 900 - 4 \times 125 = 400 \\ t1 = \frac{30 + 20}{2} = \frac{50}{2} = 25 \\ t2 = \frac{30 - 20}{2} = \frac{10}{2} = 5 \\ \\ {5}^{x} = 25 \\ x = 2 \\ \\ {5}^{x} = 5 \\ x = 1" align="absmiddle" class="latex-formula">
Ответ: 1; 2.