Решение уравнений методом вынесения множителя за скобки

1. $$7^{x+2} + 4 \cdot 7^{x+1} = 539$$$$\Rightarrow 7^{x+1}(7+4) = 539$$$$\Rightarrow 7^{x+1} \cdot 11 = 539$$$$\Rightarrow 7^{x+1} = \frac{539}{11} = 49$$$$\Rightarrow 7^{x+1} = 7^2$$$$\Rightarrow x+1 = 2$$$$\Rightarrow x = 2-1 = 1$$$$\Rightarrow x = 1$$

   Ответ:$$x = 1$$

2. $$2^{x+1} + 3 \cdot 2^{x-1} - 5 \cdot 2^x + 6 = 0$$$$\Rightarrow 2^{x-1}(2^2 + 3 - 5 \cdot 2) + 6 = 0$$$$\Rightarrow 2^{x-1}(4 + 3 - 10) = -6$$$$\Rightarrow 2^{x-1}(-3) = -6$$$$\Rightarrow 2^{x-1} = \frac{-6}{-3} = 2$$$$\Rightarrow 2^{x-1} = 2^1$$$$\Rightarrow x-1 = 1$$$$\Rightarrow x = 1+1 = 2$$$$\Rightarrow x = 2$$

   Ответ:$$x = 2$$

3. $$7^x + 7^{x+2} = 350$$$$\Rightarrow 7^x(1 + 7^2) = 350$$$$\Rightarrow 7^x(1+49) = 350$$$$\Rightarrow 7^x \cdot 50 = 350$$$$\Rightarrow 7^x = \frac{350}{50} = 7$$$$\Rightarrow 7^x = 7^1$$$$\Rightarrow x = 1$$

   Ответ:$$x = 1$$

4. $$7 \cdot 5^x - 5^{x+1} = 2 \cdot 5^3$$$$\Rightarrow 5^x(7 - 5) = 2 \cdot 5^3$$$$\Rightarrow 5^x \cdot 2 = 2 \cdot 5^3$$$$\Rightarrow 5^x = 5^3$$$$\Rightarrow x = 3$$

   Ответ:$$x = 3$$

5. $$3^{x+2} + 4 \cdot 3^{x+1} = 21$$$$\Rightarrow 3^{x+1}(3 + 4) = 21$$$$\Rightarrow 3^{x+1} \cdot 7 = 21$$$$\Rightarrow 3^{x+1} = \frac{21}{7} = 3$$$$\Rightarrow 3^{x+1} = 3^1$$$$\Rightarrow x+1 = 1$$$$\Rightarrow x = 1-1 = 0$$$$\Rightarrow x = 0$$

   Ответ:$$x = 0$$