Question

Помогите, плиизз. Хотя бы один)

Answer 1

$4^{|x| + sqrt{x^2 - 2}} - 17*2^{|x| - 2 + sqrt{x^2 - 2}} = -1\\ (2^{|x| + sqrt{x^2 - 2}})^2 - frac{17}{4}*2^{|x| + sqrt{x^2 - 2}} + 1 = 0\\ left[ (frac{17}{8})^2 - frac{225}{64} = 1 right]\\ (2^{|x| + sqrt{x^2 - 2}})^2 - 2*frac{17}{8}*2^{|x| + sqrt{x^2 - 2}} + (frac{17}{8})^2 - frac{225}{64} = 0 \\(2^{|x| + sqrt{x^2 - 2}} - frac{17}{8})^2 - frac{225}{64} = 0\\ (2^{|x| + sqrt{x^2 - 2}} - frac{17}{8} - frac{15}{8})(2^{|x| + sqrt{x^2 - 2}} - frac{17}{8} + frac{15}{8}) = 0$


$1) 2^{|x| + sqrt{x^2 - 2}} - frac{32}{8} = 0\\ 2^{|x| + sqrt{x^2 - 2}} = 2^{2}\\ |x| + sqrt{x^2 - 2} = 2\\\ x > 0; x + sqrt{x^2 - 2} = 2\\ sqrt{x^2 - 2} = 2 - x\\ x^2 - 2 = 4 - 4x + x^2\\ -4x = -6\\ x = frac{3}{2}\\\ x < 0; -x + sqrt{x^2 - 2} = 2\\ sqrt{x^2 - 2} = 2 + x\\ x^2 - 2 = 4 + 4x + x^2\\ 4x = -6\\ x = -frac{3}{2}$


$2) 2^{|x| + sqrt{x^2 - 2}} - frac{2}{8} = 0\\ 2^{|x| + sqrt{x^2 - 2}} = 2^{-2}\\ |x| + sqrt{x^2 - 2} = -2 (!)\\\ boxed{mathbb{OTBET}: x = frac{3}{2}, x = -frac{3}{2}}$