Алгебра: Решите очень вас люди добрые

Решите очень вас люди добрые

Ответы

ivan89052 15.10.2020 17:09

$2){(3 \sqrt{3} )}^{2} + {( - 3 \sqrt{3}) }^{2} - 2 = 9 \times 3 + 9 \times 3 - 2 = 27 + 27 - 2 = 52 \\ 4){( - 5 \sqrt{2} )}^{2} - {( - 2 \sqrt{5}) }^{2} - 4 = 25 \times 2 - 4 \times 5 - 4 = 50 - 20 - 4 = 26 \\ 6) \sqrt{0.87 \times 36 + 0.82 \times 36} + 1.2 = \sqrt{36(0.87 + 0.82)} + 1.2 = 6 \times 1.3 + 1.2 = 9 \\ 8) \sqrt{ \frac{72}{ {176}^{2} - {112}^{2} } } = \sqrt{ \frac{72}{(176 - 112)(176 + 112)} } = \sqrt{ \frac{72}{64 \times 288} } = \sqrt{ \frac{36}{64 \times 144} } = \frac{6}{8 \times 12} = \frac{1}{16} \\ 10) \sqrt{ \frac{ {65.5}^{2} - {15.5}^{2} }{ {13.5}^{2} - {11.5}^{2} } } = \sqrt{ \frac{(65.5 - 15.5)(65.5 + 15.5)}{(13.5 - 11.5)(13.5 + 11.5)} } = \sqrt{ \frac{50 \times 81}{2 \times 25} } = \sqrt{81} = 9$

$2) \frac{ \sqrt{11} - \sqrt{3} }{ \sqrt{11} + \sqrt{3} } - \frac{ \sqrt{11} + \sqrt{3} }{ \sqrt{11} - \sqrt{3} } = \frac{ {( \sqrt{11} - \sqrt{3} ) }^{2} - {( \sqrt{11} + \sqrt{3} )}^{2} }{( { \sqrt{11}) }^{2} - {( \sqrt{3} )}^{2} } = \frac{11 - 2 \sqrt{33} + 3 - 11 - 2 \sqrt{33} - 3}{11 - 3} = \frac{ - 4 \sqrt{33} }{8} = - \frac{ \sqrt{33} }{2}$

$4) \frac{12 + \sqrt{44} }{ 12 - \sqrt{44} } + \frac{12 - \sqrt{44} }{12 + \sqrt{44} } = \frac{ {(12 + \sqrt{44} )}^{2} + {(12 - \sqrt{44}) }^{2} }{ {(12)}^{2} - {( \sqrt{44} )}^{2} } = \frac{144 + 24 \sqrt{44} + 44 + 144 - 24 \sqrt{44} + 44 }{144 - 44} = \frac{376}{100} = 3.76$

$4 {x}^{2} + x - 1 = 0 \\ d = {b}^{2} - 4ab \\ d = 1 - 4 \times ( - 1) \times 4 \\ d = 1 + 16 = 17 \\ x1 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 1 + \sqrt{17} }{8} \\ x2 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 1 - \sqrt{17} }{8}$

${x}^{2} + 6x + 9 = 0 \\ x1 + x2 = - p \\ x1 \times x2 = q \\ \\ x1 + x2 = - 6 \\ x1 \times x2 = 9 \\ x1 = - 3 \\ x2 = - 3$

${x}^{2} + 2x + 17 = 0$

нет решения

${x}^{4} - 5 {x}^{2} + 4 = 0 \\ {x}^{2} = t \\ {t}^{2} - 5t + 4 = 0 \\ t1 = 1 \\ t2 = 4 \\ \\ {x}^{2} = 4 \\ x1 = 2 \\ x2 = - 2 \\ \\ {x}^{2} = 1 \\ x3 = 1 \\ x4 = - 1$

${(x + 3)}^{2} - |x + 3| - 30 = 0 \\ x + 3 = t \\ {t}^{2} - |t| + 30 = 0 \\ {t}^{2} - t - 30 = 0 \\ t \geqslant 0 \\ t1 = - 5 < 0 \\ t2 = 6 0 \\ t = 6 \\ \\ {t}^{2} - ( - t) - 30 = 0 \\ t < 0 \\ t1 = 5 0 \\ t2 = - 6 < 0 \\ t = - 6 \\ \\ x + 3 = 6 \\ x = 3 \\ \\ x + 3 = - 6 \\ x = - 9$

$x - 12 \sqrt{x} + 35 = 0 \\ 12 \sqrt{x} = x + 35 \\ {(12 \sqrt{x}) }^{2} = {(x + 35)}^{2} \\ 144x = {x}^{2} + 70x + 1225 \\ 144x - {x}^{2} - 70x - 1225 = 0 \\ {x}^{2} - 74x + 1225 = 0 \\ x1 = 25 \\ x2 = 49$