bantu kak, pake cara yaa

[tex]( \frac{1}{9})^{2x-3} = \sqrt{ (\frac{1}{27})^{ x^{2} -x-2} } \\ ( 3^{-2} )^{2x-3}= ( 3^{-3})^{ \frac{x^{2} -x-2}{2} } \\ 3^{6-4x}= 3^{ \frac{-3x^{2} +3x+6}{2} } \\ 6-4x=\frac{-3x^{2} +3x+6}{2} \\ 12-8x=-3x^{2} +3x+6 \\ 3x^{2} -11x+6=0 \\ (3x-2)(x-3) \\ x1 = \frac{2}{3} \\ x2= 3[/tex]