Решите уравнения:::::::

$( \frac{1}{3} )^{ \sqrt{x} }*( \frac{1}{3} )^{x}=1\\\\ ( \frac{1}{3} )^{ \sqrt{x}+x }=( \frac{1}{3} )^1\\\\ x+ \sqrt{x} =1\\\\ \left \{ {{t^2+t-1=0;\ D=1^2-4*1*(-1)=5} \atop {t= \sqrt{x} \geq 0}} \right. \\\\ \left \{ {{t= \frac{-1\pm \sqrt{5} }{2} } \atop {t= \sqrt{x} \geq 0}} \right. \\\\ \left \{ {{t= \frac{-1+ \sqrt{5} }{2} } \atop {t= \sqrt{x}}} \right. \\\\ \sqrt{x} =\frac{-1+ \sqrt{5} }{2} \\\\ x= \frac{( \sqrt{5}-1 )^2}{4} = \frac{5-2 \sqrt{5}+1 }{4}= \frac{3- \sqrt{5} }{2}$
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$4^{-x+0.5}-7*2^{-x}-4=0\\\\ 2^{2(-x+0.5)}-7*2^{-x}-4=0\\\\ 2*2^{-2x}-7*2^{-x}-4=0\\\\ \left \{ {{2t^2-7t-4=0;\ D=49+32=9^2} \atop {t=2^{-x} \ \textgreater \ 0}} \right. \\\\ \left \{ {{t= \frac{7\pm 9}{4}} \atop {t=2^{-x} \ \textgreater \ 0}} \right. \\\\ \left \{ {{t= \frac{7+9}{4}=4=2^2} \atop {t=2^{-x}}} \right. \\\\ 2^{-x}=2^2\\\\ -x=2\\\\ x=-2$