Question

помогите решить хотя бы что то

Answer 1

Ответ:

$1)(2 \frac{4}{5} - 1.8) \div \frac{4}{5} = ( \frac{14}{5} - \frac{18}{10} ) \times \frac{5}{4} = \frac{28 - 18}{10} \times \frac{5}{4} = \frac{10}{10} \times \frac{5}{4} = \frac{5}{4} = 1.25$

$2)32 \times {2}^{ - 2} \times {16}^{ - \frac{1}{2} } = 32 \times \frac{1}{ {2}^{2} } \times \frac{1}{ \sqrt{16} } = \frac{32}{4 \times 4} = \frac{32}{16} = 2$

$3) - 7 log_{12}(144) = - 7 log_{12}( {12}^{2} ) = - 7 \times 2 = - 14$

$4) {( \sqrt{6} - 3)}^{2} + 2 + 6 \sqrt{6} = 6 - 6 \sqrt{6} + 9 + 2 - 6 \sqrt{6} = 6 + 11 = 17$

$5) {4}^{3x - 2} > 256 \\ {4}^{3x - 2} > {4}^{4} \\ 3x - 2 > 4 \\ 3x >6 \\ x > 2$

$6) log_{ \frac{1}{3} }( {x}^{2} - 4x) = - 1 \\ {x}^{2} - 4x = {( \frac{1}{3} )}^{ - 1} \\ {x}^{2} - 4x = 3 \\ {x}^{2} - 4x - 3 = 0 \\ d = 16 + 12 = 28 = 4 \times 7 \\ x1 = \frac{4 + 2 \sqrt{7} }{2} = 2 + \sqrt{7} \\ x2 = 2 - \sqrt{7}$

$7) \sqrt{ \frac{4x + 27}{3} } = 11 \\ \frac{4x + 27}{3} = {11}^{2} \\ 4x + 27 = 121 \times 3 \\ 4x + 27 = 363 \\ 4x = 336 \\ x = 84$

$8) {3}^{x + 1} + {3}^{x - 1} = 270 \\ {3}^{x - 1} ( {3}^{2} + 1) = 270 \\ {3}^{x - 1} \times 10 = 270 \\ {3}^{x - 1} = 27 \\ x - 1 = 3 \\ x = 4$