Помогите пожалуйста,алгебра 8 класс,фото в закрепке

А. 1) = $\frac{(а-3b)}{5}* \frac{10ab}{ a^{2} } + \frac{6b}{a} = \frac{2ab-6 b^{2} +6 b^{2} }{a} = \frac{2ab}{a} =2b;$

A. 2) = $\frac{(x-5)(x+5)}{x+6} * \frac{1}{x(x+5)} - \frac{x+5}{x(x-6)} = \frac{x-5}{x(x+6)} - \frac{x+5}{x(x-6)} =[tex] \frac{ x^{2} -5x-6x+30 - x^{2} -5x-6x-30}{x( x^{2} -36)} = - \frac{22x}{x( x^{2} -36)} = - \frac{22}{ x^{2} -36} ;$

A.3 = $\frac{x+3}{ x^{2} +6} *( \frac{ x^{2} +3x+3x+9+ x^{2} -3x-3x+9}{(x-3)(x+3)} ) = \frac{x+3}{ x^{2} +9} * \frac{2 x^{2} +18}{(x-3)(x+3)} = \frac{x+3}{ x^{2} +9} * \$ * $\frac{2( x^{2} +9)}{(x-3)(x+3)} = \frac{2}{x-3} = \frac{2}{1.75-3} = -1.6$ ;

B.1 $= \frac{24x-16 x^{2} +9}{24x} : (\frac{3-4x}{12x} ) -2x = \frac{ (3-4x)^{2} }{24x} * \frac12{x}{3-4x} -2x = \frac{3-4x}{2} -2x = [tex] \frac{3-8x}{2} ;$