помогите сделать 2 и 4 подалуйста дам 30 балов только помогите​

Ответ:

[tex]1) \frac{2xy - y {}^{2} }{3} \times \frac{9x}{y} = y \times (2x - y) \times \frac{3x}{y} = (2x - y) \times 3x = 6x {}^{2} - 3xy[/tex]

[tex]2) \frac{a {}^{2} - 2ab }{a {}^{2} - 3a {}^{} b} \times \frac{a {}^{2}b + 3ab {}^{2} }{a {}^{3} - 2 {a}^{2} b} = \frac{a \times (a - b)}{a \times (a + 3b)}{ \times \frac{ab \times (a + 3b)}{a \times (a {}^{2} - 2ab) } } = (a - b) \times \frac{b}{a {}^{2} - 2ab } = \frac{ab - b {}^{2} }{a {}^{2} - 2ab} [/tex]

[tex]3) \frac{x {}^{2} - 25}{x {}^{2} - 6x} \times \frac{x {}^{2} - 36 }{x {}^{2} + 5x } = \frac{(x - 5) \times (x + 5)}{x {}^{6} - 6x} \times \frac{x {}^{2} - 36 }{x \times (x + 5)} = \frac{x - 5}{x {}^{6} - 6x} \times \frac{x {}^{2} - 36}{x} = \frac{(x - 5) \times (x {}^{2} - 36)}{x \times (x {}^{6} - 6x)} = \frac{x {}^{3} - 36x - 5 {x}^{2} + 180 }{ {x}^{7} - 6x {}^{2} } = \frac{ {x - 5 {x}^{2} }^{} - 36x + 180}{ {x}^{7} - 6 {x}^{2} } [/tex]

[tex]4) \frac{4 {a}^{2} - 24a + 36 }{7 {a}^{2} - 7a + 7 } \times \frac{ {a}^{3} + 1}{5 {a}^{2} - 45} = \frac{4(a {}^{2} - 6a + 9) }{7( {a}^{2} - a + 1) } \times \frac{(a + 1) \times (a {}^{2} - a + 1) }{5( {a}^{2} - 9) } = \frac{4(a - 3) {}^{2} }{7} \times \frac{a + 1}{5(a - 3) \times (a + 3)} = \frac{4(a - 3)}{7} \times \frac{a + 1}{5(a + 3)} = \frac{4(a - 3) \times (a + 1)}{35(a + 3)} = \frac{(4a - 12) \times (a + 1)}{35a + 105} = \frac{4a {}^{2} + 4 {a} - 12a - 12 }{35a + 105} = \frac{4 {a}^{2} - 8 {a} - 12 }{35a + 105} [/tex]

надеюсь помогла))

помоги мне пожалуйста с английским, буду очень благодарна :)