Помогите, нужно сделать

Решите задачу:

$2)=32 b^{2} +28b-64b-56-32 b^{2} -12b=-48b-56=48*0.5-56= \\ =-24-56=-80$
$1)= \frac{6 a^{2}+2c-6 a^{2} }{a} = \frac{2c}{a} = \frac{2*6}{25} = \frac{12}{25}$
$3)= \frac{(xy+ y^{2}*3x }{15x*(x+y)} = \frac{y*(x+y)}{5*(x+y)} = \frac{y}{5}= \frac{7}{5}$
$4)= \frac{(a+x)*a}{a*(ax+ x^{2})} = \frac{a+x}{x*(a+x)} = \frac{1}{x} = \frac{1}{15}$
$5)= \frac{ x^{2} *( x^{2} -49y)}{x*( x^{2} -7xy)} = \frac{ x^{2} *(x-7y)(x+7y)}{x*( x^{2} -7xy)} = \frac{x*(x-7y)(x+7y)}{x(x-7y)}=x+7y=2+7 \sqrt{2} +49-7 \sqrt{2} =51$