Question

$5^x +5^-^x = 7/4$

Answer 1

ответ:

$\displaystyle x \notin R$

решение:

$\displaystyle 5^x+5^-^x=\frac{7}{4}\\ \\5^x+\frac{1}{5^x}=\frac{7}{4}\\ \\t+\frac{1}{t}=\frac{7}4}\\ \\t+\frac{1}{t}-\frac{7}{4}=0\\ \\\frac{4t^2+4-7t}{4t}=0\\ \\4t^2+4-7t=0\\ 4t^2-7t+4=0\\ \\t=\frac{-(-7)б\sqrt{(-7)^2-4*4*4} }{2*4}=\frac{7б\sqrt{(-7)^2-4*4*4} }{2*4}=\frac{7б\sqrt{49-4*4*4} }{2*4}=\\\frac{7б\sqrt{49-64} }{2*4}=\frac{7б\sqrt{49-64} }{8}=\frac{7б\sqrt{-15} }{8}\\ \\t \notin R\\ x \notin R$

формула:

$\displaystyle a^-^n=\frac{1}{a^n}$