Question

!СРОЧНО! ДАЮ 50 БАЛЛОВ!

Answer 1

Решение в фото:

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Answer 2

Объяснение:

1)

$\frac{200 {x}^{4} {y}^{3} - 55 { {x}^{3} {y}^{2} } }{5 {x}^{3} {y}^{2} } + 11 = \frac{5 {x}^{3} {y}^{2} (40xy - 11 }{5 {x}^{3} {y}^{2} } + 11 = 40xy - 11 + 11 = 40xy$

тождество доказано.

2.

$\frac{11}{10} a - \frac{12.1 {a}^{3} {b}^{2} - 44 {a}^{2} {b}^{2} }{11 {a}^{2} {b}^{2} } = \frac{11}{10} a - \frac{ {a}^{2} {b}^{2}(12.1a - 44) }{11 {a}^{2} {b}^{2} } = \frac{11}{10} a - \frac{12.1a - 44}{11} = \frac{11}{10} a - \frac{ \frac{121}{10}a - 44 }{11} = \frac{11}{10} a - \frac{ \frac{121a - 440}{10} }{11} = \frac{11}{10} a - \frac{121a - 440}{110} = \frac{11}{10} a - \frac{11(11 - 40)}{110} = \frac{11}{10} a - \frac{11a - 40}{10} = \frac{11a - (11a - 40)}{10} = \frac{11a - 11a + 40}{10} = \frac{40}{10} = 4$

тождество доказано