Question

$3^{x^2-x}+3^{3+x-x^2} =12$
решить показательное уравнение

Answer 1

a^-n = 1/a^n

a^m * a^n = a^(m+n)

3^(x^2 - x) + 3^(3 + x - x^2) = 12

3^(x^2 - x) + 3^3*3^(x - x^2) = 12

3^(x^2 - x) + 27/3^(x^2 - x) = 12

3^(x^2 - x) = t > 0

t + 27/t - 12 = 0

t^2 - 12t + 27 = 0

D = 12^2 - 4*27 = 144 - 108 = 36

t12 = (12 +- 6)/2 = 3    9

t1 = 3

3^(x^2 - x) = 3

x^2 - x = 1

x^2 - x - 1 = 0

D = 1 + 4 = 5

x12 = (1 +- √5)/2

t2= 9

3^(x^2 - x) = 3^2

x^2 - x = 2

x^2 - x - 2 = 0

D = 1 + 8 = 9

x34 = (1 +- 3)/2 = -1   2

ответ -1, 2,  (1 + √5)/2, (1 - √5)/2

Answer 2

Ответ:

$x_{1} = -1 , x_{2} = \frac{1 - \sqrt{5} }{2} , x_{3} = \frac{1 + \sqrt{5} }{2} , x_{4} = 2$

Объяснение:

$x^{x^{2} -x} + 3^{3+x -x^{2} } = 12$

$\frac{1}{3^{-x^{2}+ x } } + 3^{3} * 3^{x - x^{2} } = 12$

$\frac{1}{3 - x^{2} + x} + 27 * 3^{-x^{2}+ x } = 12$

$\frac{1}{t} + 27t = 12$

$t = \frac{1}{9} \\t = \frac{1}{3}$

$3^{-x^{2} + x } = \frac{1}{9} \\3^{-x^{2} + x } = \frac{1}{3}$

$x = -1\\x = 2\\x = \frac{1 + \sqrt{5} }{2} \\x = \frac{1 - \sqrt{5} }{2}$

Ответ: $x_{1} = -1 , x_{2} = \frac{1 - \sqrt{5} }{2} , x_{3} = \frac{1 + \sqrt{5} }{2} , x_{4} = 2$