NOMOR 1.
[tex]{(9)}^{2x + 6} = {(3)}^{3x - 5} [/tex]
[tex]{({3}^{2})}^{2x + 6} = {(3)}^{3x - 5} [/tex]
[tex]{(3)}^{(2)(2x + 6)} = {(3)}^{3x - 5} [/tex]
[tex]{(3)}^{4x + 12} = {(3)}^{3x - 5} [/tex]
Sehingga, nilai x adalah.
[tex]4x + 12 = 3x - 5[/tex]
[tex]4x - 3x = - 12 - 5[/tex]
[tex]x = - 17[/tex]
NOMOR 2.
[tex]\frac{6}{4 \sqrt{2} + \sqrt{5}} = \frac{6}{4 \sqrt{2} + \sqrt{5}} \times \frac{4 \sqrt{2} - \sqrt{5}}{4 \sqrt{2} - \sqrt{5}} [/tex] [tex]= \frac{(6)(4 \sqrt{2} - \sqrt{5})}{(4 \sqrt{2} + \sqrt{5})(4 \sqrt{2} - \sqrt{5})} [/tex]
[tex]= \frac{24 \sqrt{2} - 6 \sqrt{5}}{16 \sqrt{4} - 4 \sqrt{10} + 4 \sqrt{10} - \sqrt{25} } [/tex]
[tex]= \frac{24 \sqrt{2} - 6 \sqrt{5}}{16 \sqrt{4} - \sqrt{25}} [/tex]
[tex]= \frac{24 \sqrt{2} - 6 \sqrt{5}}{16 \times 2 - 5} = \frac{24 \sqrt{2} - 6 \sqrt{5}}{32 - 5} = \frac{24 \sqrt{2} - 6 \sqrt{5}}{27} = \frac{8 \sqrt{2} - 2 \sqrt{5}}{9} [/tex]