1) 2^2-x=4 2) 8=4^1/2x+1 3) (12/17)^x/2 +1 =(5/20)^x/2 +1 4) 2^2-x-2^2x-1=1

Решите задачу:

$2 {}^{2 - x} = 2 {}^{2} \\ 2 - x = 2 \\ - x = 2 - 2 \\ - x = 0 \\ x = 0$

$2 {}^{3} = 2 {}^{ \frac{2}{2x + 1} } \\ 3 = \frac{2}{2x + 1} \\ 3(2x + 1) = 2 \\ 6x + 3 = 2 \\ 6x = - 1 \\ x = - \frac{1}{6}$
$(2 - 1) \times 2 {}^{2x - 1} = 1 \\ 1 \times 2 {}^{2x - 1} = 1 \\ 2 {}^{2x - 1} = 1 \\ 2 {}^{2x - 1} = 2 {}^{0} \\ 2x - 1 = 0 \\ 2x = 1 \\ x = \frac{1}{2}$

$( \frac{12}{17} ) {}^{ \frac{x}{2} + 1 } = ( \frac{1}{4}) {}^{ \frac{x}{2} + 1 } \\ ( \frac{ \frac{12}{17} }{ \frac{1}{4} } ) {}^{ \frac{x}{2} + 1} = 1 \\ ( \frac{48}{17}) {}^{ \frac{x}{2} + 1 } = ( \frac{48}{17} ) {}^{0} \\ \frac{x}{2} + 1 = 0 \\ x + 2 = 0 \\ x = - 2$