1. Решите уравнение: 1) 6ˣ + 6ˣ⁻¹ - 6ˣ⁻² = 7ˣ - 8 * 7ˣ⁻²; 2) 5ˣ - 2 * 5ˣ⁻¹ = 3ˣ⁺¹ - 2 * 3ˣ⁻²; 3) 2√ˣ⁺¹ - 3√ˣ = 3√ˣ⁻¹ - 2√ˣ. 2. Решите уравнение: 1) 8²/ˣ - 2⁽²ˣ⁺³⁾/ˣ - 32 = 0; 2) 5√ˣ⁻² - 5¹⁻√ˣ⁻² - 4 = 0;

Решим данные уравнения.

1.

1) $$6^x + 6^{x-1} - 6^{x-2} = 7^x - 8 \cdot 7^{x-2};$$

$$6^x + \frac{6^x}{6} - \frac{6^x}{6^2} = 7^x - 8 \cdot \frac{7^x}{7^2};$$

$$6^x(1 + \frac{1}{6} - \frac{1}{36}) = 7^x(1 - \frac{8}{49});$$

$$6^x(\frac{36+6-1}{36}) = 7^x(\frac{49-8}{49});$$

$$6^x(\frac{41}{36}) = 7^x(\frac{41}{49});$$

$$\frac{6^x}{7^x} = \frac{41}{49} : \frac{41}{36};$$

$$(\frac{6}{7})^x = \frac{36}{49};$$

$$(\frac{6}{7})^x = (\frac{6}{7})^2;$$

$$x = 2.$$

Ответ: 2

2) $$5^x - 2 \cdot 5^{x-1} = 3^{x+1} - 2 \cdot 3^{x-2};$$

$$5^x - 2 \cdot \frac{5^x}{5} = 3 \cdot 3^x - 2 \cdot \frac{3^x}{3^2};$$

$$5^x(1 - \frac{2}{5}) = 3^x(3 - \frac{2}{9});$$

$$5^x(\frac{5-2}{5}) = 3^x(\frac{27-2}{9});$$

$$5^x(\frac{3}{5}) = 3^x(\frac{25}{9});$$

$$\frac{5^x}{3^x} = \frac{25}{9} : \frac{3}{5};$$

$$(\frac{5}{3})^x = \frac{25}{9} \cdot \frac{5}{3};$$

$$(\frac{5}{3})^x = \frac{125}{27};$$

$$(\frac{5}{3})^x = (\frac{5}{3})^3;$$

$$x = 3.$$

Ответ: 3

3) $$2^{\sqrt{x}+1} - 3^{\sqrt{x}} = 3^{\sqrt{x}-1} - 2^{\sqrt{x}};$$

$$2^{\sqrt{x}+1} + 2^{\sqrt{x}} = 3^{\sqrt{x}-1} + 3^{\sqrt{x}};$$

$$2^{\sqrt{x}} \cdot 2 + 2^{\sqrt{x}} = 3^{\sqrt{x}} + \frac{3^{\sqrt{x}}}{3};$$

$$2^{\sqrt{x}}(2+1) = 3^{\sqrt{x}}(1 + \frac{1}{3});$$

$$2^{\sqrt{x}} \cdot 3 = 3^{\sqrt{x}} \cdot \frac{4}{3};$$

$$\frac{2^{\sqrt{x}}}{3^{\sqrt{x}}} = \frac{4}{3} : 3;$$

$$(\frac{2}{3})^{\sqrt{x}} = \frac{4}{3} \cdot \frac{1}{3};$$

$$(\frac{2}{3})^{\sqrt{x}} = \frac{4}{9};$$

$$(\frac{2}{3})^{\sqrt{x}} = (\frac{2}{3})^2;$$

$$\sqrt{x} = 2;$$

$$x = 4.$$

Ответ: 4

2.

1) $$8^{\frac{2}{x}} - 2^{\frac{2x+3}{x}} - 32 = 0;$$

$$(2^3)^{\frac{2}{x}} - 2^{\frac{2x+3}{x}} - 32 = 0;$$

$$2^{\frac{6}{x}} - 2^{\frac{2x}{x} + \frac{3}{x}} - 32 = 0;$$

$$2^{\frac{6}{x}} - 2^{2 + \frac{3}{x}} - 32 = 0;$$

$$2^{\frac{6}{x}} - 2^2 \cdot 2^{\frac{3}{x}} - 32 = 0;$$

$$2^{\frac{6}{x}} - 4 \cdot 2^{\frac{3}{x}} - 32 = 0;$$

Пусть $$2^{\frac{3}{x}} = t$$, тогда $$2^{\frac{6}{x}} = (2^{\frac{3}{x}})^2 = t^2$$.

$$t^2 - 4t - 32 = 0;$$

$$D = (-4)^2 - 4 \cdot 1 \cdot (-32) = 16 + 128 = 144;$$

$$t_1 = \frac{4 + \sqrt{144}}{2 \cdot 1} = \frac{4 + 12}{2} = \frac{16}{2} = 8;$$

$$t_2 = \frac{4 - \sqrt{144}}{2 \cdot 1} = \frac{4 - 12}{2} = \frac{-8}{2} = -4.$$

Вернемся к замене:

$$2^{\frac{3}{x}} = 8;$$

$$2^{\frac{3}{x}} = 2^3;$$

$$\frac{3}{x} = 3;$$

$$x = 1.$$

$$2^{\frac{3}{x}} = -4$$ - решения нет, т.к. показательная функция всегда больше нуля.

Ответ: 1

2) $$5^{\sqrt{x-2}} - 5^{1-\sqrt{x-2}} - 4 = 0;$$

$$5^{\sqrt{x-2}} - \frac{5}{5^{\sqrt{x-2}}} - 4 = 0;$$

Пусть $$5^{\sqrt{x-2}} = t$$, тогда

$$t - \frac{5}{t} - 4 = 0;$$

$$t^2 - 5 - 4t = 0;$$

$$t^2 - 4t - 5 = 0;$$

$$D = (-4)^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36;$$

$$t_1 = \frac{4 + \sqrt{36}}{2 \cdot 1} = \frac{4 + 6}{2} = \frac{10}{2} = 5;$$

$$t_2 = \frac{4 - \sqrt{36}}{2 \cdot 1} = \frac{4 - 6}{2} = \frac{-2}{2} = -1.$$

Вернемся к замене:

$$5^{\sqrt{x-2}} = 5;$$

$$5^{\sqrt{x-2}} = 5^1;$$

$$\sqrt{x-2} = 1;$$

$$x - 2 = 1;$$

$$x = 3.$$

$$5^{\sqrt{x-2}} = -1$$ - решения нет, т.к. показательная функция всегда больше нуля.

Ответ: 3