kak tolong dijawab dong kak

(1/3)^(2x²-7x+6) = 1
3^[-1(2x² - 7x + 6)] = 3^0

-1 (2x²-7x + 6) = 0
-2x² + 7x - 6 = 0
2x² - 7x + 6 = 0
(x-2) (2x-3) = 0
x1 = 2
x2 = 3/2

(1/3) 2x"- 1x + 6 -1 =0
2x"+ (1/3) -1x + 5=0
2x" + 1/3+ 5- 1x=0
2x" + 6/3- 1x=0
2x" + 2 -1x=0
2x" - x + 2= 0

menggunakan rumus ax" + bx + c =0
keterangan :
(") pangkat 2
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