17 1) x²-9=0 2) 0,5⁶⁻²ˣ=32 3) log₇(2x+5)=1

• $$x^2 - 9 = 0$$

  $$x^2 = 9$$

  $$x = \pm 3$$

• $$0,5^{6-2x} = 32$$

  $$(2^{-1})^{6-2x} = 2^5$$

  $$2^{2x-6} = 2^5$$

  $$2x - 6 = 5$$

  $$2x = 11$$

  $$x = 5,5$$

• $$\log_7(2x+5) = 1$$

  $$2x+5 = 7^1$$

  $$2x+5 = 7$$

  $$2x = 2$$

  $$x = 1$$

Ответ: 1) ±3; 2) 5,5; 3) 1