Задание 271

$$\left(\frac{1}{7}\right)^{2x+1}:\frac{1}{49}=49^x\cdot 7$$

$$\left(\frac{1}{7}\right)^{2x+1-2}=7^{2x+1}$$

$$\left(\frac{1}{7}\right)^{2x-1}=7^{2x+1}$$

$$7^{1-2x}=7^{2x+1}$$

$$1-2x=2x+1$$

$$2x+2x=1-1$$

$$4x=0$$

$$x=0$$