Question

помогите, с меня 30 баллов ​

Answer 1

▪3

$79.04 \div ( - 2.6) - 18.93 = \frac{7904}{100} \div ( - \frac{26}{10} ) - 18.93 = - \frac{7904 \times 10}{100 \times 26} - 18.93 = - \frac{304}{10} - 18.93 = - 30.4 - 18.93 = - 49.33$

▪4

$\sqrt{64} d - \frac{1}{7} \sqrt{49} d = 8d - \frac{1}{7} \times 7d = 8d - d = 7d$

▪5.1

$7 {x}^{2} - 21 = 0 \\ 7 {x}^{2} = 21 \\ {x}^{2} = 21 \div 7 \\ {x}^{2} = 3 \\ x = ± \sqrt{3}$

▪5.2

$5 {x}^{2} + 9x = 0 \\ x(5x + 9) = 0 \\ x_{1} = 0 \: \: \: \: \: \: \: 5x + 9 = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 5x = - 9 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = - \frac{9}{5} = - 1 \frac{4}{5} = - 1.8$

▪5.3

${x}^{2} + x - 49 = 0 \\ d = 1 - 4 \times ( - 49) = 1 + 196 = 197 \\ x_{1,2} = \frac{ - 1 ± \sqrt{197} }{2}$

▪7.

$5 {x}^{4} + 3 {x}^{2} - 2 = 0 \\ {x}^{2} = y \\ 5 {y}^{2} + 3y - 2 = 0 \\ d = 9 - 4 \times 5 \times ( - 2) = 9 + 40 = 49 \\ y_{1} = \frac{ - 3 + \sqrt{49} }{2 \times 5} = \frac{ - 3 + 7}{10} = \frac{4}{10} = 0.4 \\ y_{2} = \frac{ - 3 - \sqrt{49} }{2 \times 5} = \frac{ - 3 - 7}{10} = \frac{ - 10}{10} = - 1 \\ proizvedem \: obratnuju \: zamenu \\ {x}^{2} = y_{1} \: \: \: \: \: \: \: \: {x}^{2} = y_{2} \\ {x}^{2} = 0.4 \: \: \: \: \: \: {x}^{2} = - 1 \: ne \: podhodit\\ x_{1} = ± \sqrt{0.4} \: \: \: \:$

▪9.

$( \frac{a - 2}{a + 2} + \frac{a + 2}{a - 2} ) \div \frac{12 {a}^{2} }{4 - {a}^{2} } = \frac{ {(a - 2)}^{2} + {(a + 2)}^{2} }{ {a}^{2} - 4 } \times ( - \frac{ {a}^{2} - 4}{12 {a}^{2} }) = - \frac{ {a}^{2} - 4a + 4 + {a}^{2} + 4a + 4 }{12 {a}^{2} } = - \frac{2 {a}^{2} + 8 }{12 {a}^{2} } = - \frac{2( {a}^{2} + 4)}{12 {a}^{2} } = - \frac{ {a}^{2} + 4}{6 {a}^{2} }$