Question

8^x x 2^x+2 x 4^x= 2 ^3x +11
aflati termenul necunoscut ! Va rog!​

Answer 1

Răspuns:

$8x \times 2x + 2 \times 4x = 2 \times 3x + 11 = \geqslant16x {}^{2} + 8x = 6x + 11 = > 16x {}^{2} + 8x - 6x = 11 = > 16x {}^{2} + 2x - 11 = 0 = > \\ a = 16 \\ b = 2 \\ c = - 11 \\ delta = 2 {}^{2} - 4 \times 16 \times ( - 11) = > 4 - 4 \times ( - 17) = > 4 - 684 = > - 680 \\ x1si2 = \frac{ - b + - \sqrt{delta} }{2a} = \frac{ - 2 + - \sqrt{ - 680} }{2 \times 16} = \frac{ - 2 + - ( - 2 \sqrt{170)} }{32} = > x1 = \frac{ - 2 + ( - 2 \sqrt{170)} }{32} = > \frac{ - 2 - 2 \sqrt{170} }{32} = > \frac{ - 2( - 1 \sqrt{170)} }{32} = > \frac{ - 1( - 1 \sqrt{170)} }{32} \\ x2 = \frac{ - 2 - 2 \sqrt{170} }{32} = > - \frac{1 + \sqrt{170} }{16}$

Coroană!!