Question

решите
2^x - 8 * 2^-x = 7

Answer 1

${2}^{x} - 8 times {2}^{ - x} = 7 \ {2}^{x} - frac{8}{ {2}^{x} } = 7 : : : |zamenim : : {2}^{x} = y \ y - frac{8}{y} = 7 : : : | times y \ {y}^{2} - 8 = 7y \ {y}^{2} - 7y - 8= 0 \ d = {b}^{2} - 4ac = {7}^{2} - 4 times ( - 8) = 49 +32 = 81 \ y1 = frac{ - b + sqrt{d} }{2a} = frac{7 + sqrt{81} }{2} = frac{7 + 9}{2} = frac{16}{2} = 8 \ y2 = frac{ - b - sqrt{d} }{2a} = frac{7 - sqrt{81} }{2} = frac{7 - 9}{2} = frac{ - 2}{2} = - 1 \ - - - - - - - - \ {2}^{x} = y1 \ {2}^{x} = 8 \ {2}^{x} = {2}^{3} \ x1 = 3 \ - - - - - - \ {2}^{x} = y2 \ {2}^{x} = - 1$
Ответ: х = 3

Answer 2

$2^x-8* 2^{-x}=7 \ 2^x-8* frac{1}{2x}=7 \ 2^x=t,t textgreater 0 \ t-8* frac{1}{t} =7 \ frac{t^2-8-7t}{t} =0 \ t^2-7t-8=0 \ a-b+c=0 \ t=-1 \ t=8 \ 2^x=8 \ x=3 \ 2^x=-1$
x∈∅
ответ 3