Question

Помогите пожалуйста

Answer 1

$9 ^{3-x}= \sqrt{3} \\\\(3 ^{2} ) ^{3-x} =3 ^{ \frac{1}{2} }\\\\3 ^{6-2x}=3 ^{0,5}\\\\6-2x=0,5\\\\2x=6-0,5\\\\2x=5,5\\\\x=2,75\\\\\\ \frac{1}{2}( \frac{9}{25}) ^{x} = \frac{27}{250} \\\\( \frac{9}{25}) ^{x} = \frac{27}{125} \\\\( \frac{3}{5} ) ^{2x} =( \frac{3}{5}) ^{3}\\\\2x= 3\\\\x=1,5\\\\\\0,3 ^{ x^{2} -x} =1\\\\0,3 ^{ x^{2} -x}=0,3^{o} \\\\ x^{2} -x=0\\\\x(x-1)=0\\\\ x_{1} =0\\\\x-1=0\\\\ x_{2}=1$