Помогите, то что находится в кругах

1-a)
$\frac{60 {x}^{3} }{100 {x}^{4} } = \frac{3}{5x}$
b)
$(2x - 6) \times \frac{ {x}^{2} }{x - 3} = \frac{ {x}^{2}(2x - 6) }{x - 3} = \frac{2 {x}^{3} - 6 {x}^{2} }{x - 3} = \frac{2 {x}^{2}(x - 3) }{x - 3} = 2 {x}^{2}$
в)
${ ( -\frac{3a}{ {b}^{2} } })^{2} = \frac{9 {a}^{2} }{ {b}^{4} }$
г)
$\frac{ {y}^{2} - y}{2xy} \times \frac{2x}{ {y}^{2} - 1 } = \frac{y(y - 1) \times 2 x}{2xy(y - 1)(y + 1)} = \frac{1}{(y + 1)}$
2-а
$- \frac{4a}{5 {c}^{2} } \times - \frac{10 {c}^{4} }{a} = - \frac{8 a{c}^{2} }{a} = - \frac{8 {c}^{2} }{a}$
б)
$\frac{ {m}^{2} - 4 {n}^{2} }{ {m}^{2} } \times \frac{1}{4m - 8n} = \frac{(m - 2n)(m + 2n)}{4 {m}^{2} (m - 2n) } = \frac{m + 2n}{4 {m}^{2} }$
в)
$\frac{2x - 1}{ {x}^{2} - 6x + 9 } \times \frac{ {x}^{2} - 3x }{1 - 2x} = \frac{2x + 1}{ {(x - 3)}^{2} } \times \frac{x(x - 3)}{1 - 2x} = \frac{ x(2 x + 1)}{(x - 3)(1 - 2 x)}$
г)
$\frac{2a - 4}{ {a}^{2} + 4} \div \frac{ {(a - 2)}^{2} }{( {a}^{2} - 4)( {a}^{2} + 4) } = \frac{2(a - 2)}{ {a}^{2} + 4} \times \frac{( {a}^{2} - 4)( {a}^{2} + 4) }{ {(a - 2)}^{2} } = \frac{2( {a }^{2} - 4)}{a - 2} = \frac{2 {a }^{2} - 8}{a - 2}$