Question

Помогите пожалуйста решить

Answer 1

$1.$

$a) \: \: \: \: \: 20 - 18 .6 \div (6 \frac{11}{15} - 4 \frac{3}{20} ) = 1.4 \div ( \frac{101}{15} - \frac{83}{20} ) = 1.4 \div \frac{404 - 249}{60} = \frac{14}{10} \div \frac{155}{60} = \frac{14}{10} \times \frac{60}{155} = \frac{84}{155}$

$b) \: \: \: \: \: 5 \frac{3}{7} - (2 \frac{1}{2} + 1 \frac{1}{3} ) \div \frac{1}{6} = \frac{38}{7} - ( \frac{5}{2} + \frac{4}{3} ) \times 6 = \frac{38}{7} - \frac{15 + 8}{6} \times 6 = \frac{38}{7} - \frac{23}{6} \times 6 = \frac{38}{7} - 23 = \frac{38 - 161}{7} = - \frac{123}{7} = 17 \frac{4}{7}$

$2.$

${(x - y)}^{2} - x(x - 2y) = {x}^{2} - 2xy + {y}^{2} - {x}^{2} + 2xy = {y}^{2}$

$3.$

$a) \: \: \: \: \: {3x}^{2} + 7x - 6 = 0 \\ D = {7}^{2} - 4 \times 3 \times ( - 6) = 49 + 72 = 121 \\ x1x2 = \frac{ - 7± \sqrt{121} }{2 \times 3} = \frac{ - 7±11}{6} \\ x1x2 = - 3 \: ; \: \frac{2}{3}$

$b) \: \: \: \: \: \frac{x + 9}{3} - \frac{x}{5} = 1 \\ \frac{5x + 45 - 3x}{15} = 1 \\ 2x + 45 = 15 \\ 2x = - 30 \\ x = - 15$

$c) \: \: \: \: \: \frac{x}{x + 2} + \frac{x + 2}{x - 2} = \frac{8}{ {x}^{2} - 4 } \\ \frac{ {x}^{2} - 2x + {x}^{2} + 4x + 4 }{(x + 2)(x - 2)} = \frac{8}{ {x}^{2} - 4} \\ 2 {x}^{2} + 2x + 4 = 8 \\ 2 {x}^{2} + 2x - 4 = 0 \\ {x}^{2} + x - 2 = 0 \\ x1 + x2 = - 1 \\ x1 \times x2 = - 2 \\ x1 = - 2 \\ x2 = 1$