Question

Решите показательные неравенства

Answer 1

1 вариант

1.

$1) {4}^{3x - 17} = {4}^{3} \\ 3x - 17 = 3 \\ 3x = 20 \\ x = \frac{20}{3}$

$2) {5}^{x + 1} + {5}^{x - 1} = 130 \\ {5}^{x - 1} ( {5}^{2} + 1) = 130 \\ {5}^{x - 1} \times 26 = 130 \\ {5}^{x - 1} = 5 \\ x - 1 = 1 \\ x = 2$

2.

${6}^{ {x}^{2} + x - 4 } \leqslant {6}^{2} \\ {x}^{2} + x - 4 \leqslant 2 \\ {x}^{2} + x - 6 \leqslant 0 \\ d = 1 + 24 = 25 \\ x1 = ( - 1 + 5) \div 2 = 2 \\ x2 = - 3$

ответ: х принадлежит [-3;2].

2 вариант

1.

$1) {3}^{5x + 12} = {3}^{4} \\ 5x + 12 = 4 \\ 5x = - 8 \\ x = - \frac{8}{5}$

$2) {2}^{x + 3} - {2}^{x - 1} = 60 \\ {2}^{x - 1} ( {2}^{4} - 1) = 60 \\ {2}^{x - 1} \times 15 = 60 \\ {2}^{x - 1} = 4 \\ x - 1 = 2 \\ x = 3$

2.

${7}^{ {x}^{2} - 2x - 7} \geqslant 7 \\ {x}^{2} - 2x - 7 \geqslant 1 \\ {x}^{2} - 2x - 8 \geqslant 0 \\ d = 4 + 32 = 36 \\ x1 = 4 \\ x2 = - 2$

ответ: х принадлежит

$( - \infty; - 2]U[4 ;+ \infty )$

3 вариант

1.

$1) {6}^{2x + 11} = {6}^{3} \\ 2x + 11 = 3 \\ 2x = - 8 \\ x = - 4$

$2) {4}^{x + 2} + {4}^{x - 1} = 260 \\ {4}^{x - 1} ( {4}^{3} + 1) = 260 \\ {4}^{x - 1} \times 65 = 260 \\ {4}^{x - 1} = 4 \\ x = 2$

2.

${5}^{ {x }^{2} - 3x - 2} \leqslant 25 \\ {x}^{2} - 3x - 2 \leqslant 2 \\ {x}^{2} - 3x - 4 \leqslant 0 \\ d = 9 + 16 = 25 \\ x1 = 4 \\ x2 = - 1$

ответ: х принадлежит [-1;4]

4 вариант.

1.

$1) {5}^{4x + 7} = {5}^{3} \\ 4x + 7 = 3 \\ 4x = - 4 \\ x = - 1$

$2) {3}^{x + 1} - {3}^{x - 1} = 72 \\ {3}^{x - 1} ( {3}^{2} - 1) = 72 \\ {3}^{x - 1} = 9 \\ x - 1 = 2 \\ x = 3$

2.

${4}^{ {x}^{2} + 5x - 12 } \geqslant 16 \\ {x}^{2} + 5x - 12 \geqslant 2 \\ {x}^{2} + 5x - 14 \geqslant 0 \\ d = 25 + 56 = 81 \\ x1 = ( - 5 + 9) \div 2 = 2 \\ x2 = - 7$

ответ: х принадлежит

$( - \infty; - 7]U[2 ;+ \infty )$