ОДЗ :
x² - 2 ≥ 0
x ∈ (- ∞ ; - √2] ∪ [√2 ; + ∞)
0\\\\2t^{2}-5t-12=0\\\\D=(-5)^{2}-4*2*(-12)=25+96=121=11^{2}\\\\t_{1}=\frac{5+11}{4}=4\\\\t_{2}=\frac{5-11}{4}=-\frac{3}{2}<0" alt="4^{\sqrt{x^{2}-2} +x}-5*2^{\sqrt{x^{2}-2 }+x-1 }=6\\\\4^{\sqrt{x^{2}-2 }+x } -\frac{5}{2}*2^{\sqrt{x^{2}-2 }+x }-6=0\\\\2*4^{\sqrt{x^{2}-2 }+x }-5*2^{\sqrt{x^{2}-2 }+x }-12=0\\\\2^{\sqrt{x^{2}-2 } +x}=t,t>0\\\\2t^{2}-5t-12=0\\\\D=(-5)^{2}-4*2*(-12)=25+96=121=11^{2}\\\\t_{1}=\frac{5+11}{4}=4\\\\t_{2}=\frac{5-11}{4}=-\frac{3}{2}<0" align="absmiddle" class="latex-formula">
Ответ : 1,5