Question

9 класс. Нужно полное и подробное решение заданий.

Answer 1

1)
$(5 \frac{2}{3} - 5 \frac{13}{15} \div 2.2) \times 1 \frac{1}{3} = ( \frac{17}{3} - \frac{88}{15} \div \frac{22}{10} ) \times \frac{4}{3} = ( \frac{17}{3} - \frac{88}{15} \times \frac{10}{22} ) \times \frac{4}{3} = ( \frac{17}{3} - \frac{4 \times2}{3 \times 1} ) \times \frac{4}{3} = ( \frac{17}{3} - \frac{8}{3} ) \times \frac{4}{3} = \frac{9}{3} \times \frac{4}{3} = 3 \times \frac{4}{3} = 4$
Ответ: в)

2)
$\frac{2 {a}^{6} {b}^{2} }{3 {c}^{5} } \times \frac{9 {b}^{3} {c}^{3} }{8 {a}^{4} } \div \frac{3 {a}^{2} {b}^{5} }{4 {c}^{2} } = \frac{18{a}^{6 - 4} {b}^{2 + 3} }{24 {c}^{5 - 3} } \times \frac{4 {c}^{2} }{3 {a}^{2} {b}^{5} } = \frac{3 {a}^{2} {b}^{5} }{4 {c}^{2} } \times \frac{4 {c}^{2} }{3 {a}^{2} {b}^{5} } = 1$
Ответ: г)

3)
$(x - 2)(x + 3) = 24 \\ {x}^{2} - 2x + 3x - 6 = 24 \\ {x}^{2} + x - 6 - 24 = 0 \\ {x}^{2} + x - 30 = 0 \\ d = {b}^{2} - 4ac = 1 - 4 \times 1 \times ( - 30) = 121 \\ x1 = \frac{ - 1 + 11}{2} = \frac{10}{2} = 5 \\ x2 = \frac{ - 1 - 11}{2} = \frac{ - 12}{2} = - 6$
Ответ: а)