Вычислить Помогите!!!!!!!!!!! Прошу

A) $\frac{( 16^{log_{4}( \sqrt{5}-1 ) } + 9^{log_{3}( \sqrt{5}+1 ) } )log_{3}4 }{log _{3} 64} = \frac{( (4^{log_{4}( \sqrt{5}-1 ) })^{2} + (3^{log_{3}( \sqrt{5}+1 ) })^{2} )log_{3}4 }{3log _{3} 4} = \frac{( ( \sqrt{5}-1 )^{2} + ( \sqrt{5}+1 )^{2} ) }{3} = \frac{ 25-2 \sqrt{5}+1 + 25+2 \sqrt{5}+1 }{3} = \frac{ 52 }{3} = 17\frac{1}{3}$

б) $(\frac{1}{9})^{x} + 8 (\frac{1}{3})^{x} - 9 \leq 0$
$3^{-2x} + 8*3^{-x} - 9 \leq 0$
Пусть $3^{x} = t$ при 0 " alt=" t > 0 " align="absmiddle" class="latex-formula">
$\frac{1}{t^{2}}+ \frac{8}{t} - 9 = 0$
$\frac{-9t^{2} + 8t + 1}{ t^{2} } = 0$
$-9t^{2} + 9t - t + 1 = 0$
$-9t(t - 1) - 1(t - 1)} = 0$
$-(9t + 1)(t - 1) = 0$
$t = -\frac{1}{9}; 1$
0" alt="t > 0" align="absmiddle" class="latex-formula"> поэтому $t = 1$
ВОЗ: $3^{x}=1$
$x=0$

в) $log_{3}x+4log_{9}x = 9$
$log_{3}x+2log_{3}x = 9$
$3log_{3}x = 9$
$log_{3}x = 3$
$x = 27$