Question

Умоляю решите с решением!!!!!​

Answer 1

$1)\frac{b^{2}}{2a}*\frac{6a}{ab-b^{2}}=\frac{b^{2}}{2a}*\frac{6a}{b(a-b)}=\frac{3b}{a-b}\\\\2)\frac{ab+ac}{bc}:\frac{ab-ac}{bc^{2}}=\frac{a(b+c)}{bc}*\frac{bc^{2}}{a(b-c)}=\frac{c(b+c)}{b-c}\\\\3)\frac{x}{1-9x^{2}}*\frac{3x-1}{5x^{2}}=\frac{x}{(1-3x)(1+3x)}*\frac{3x-1}{5x^{2}}=-\frac{1}{5x(1+3x)}$

$4)\frac{x^{2}-9 }{x^{2}+4x+4}:\frac{3-x}{3x+6}*\frac{x}{x+3}=\frac{(x-3)(x+3)}{(x+2)^{2}}*\frac{3(x+2)}{3-x}*\frac{x}{x+3}=-\frac{3x}{x+2}$

Answer 2

1). (b^2/2a)*(6a/(ab-b^2)=(b^2/2a)*(6a/(b*(a-b)))=b*(3/(a-b))

2). ((ab+ac)/bc)/((ab-ac)/bc^2)=((a(b+c))/bc)*(bc^2/(a*(b-c)))=(b+c)*(bc/(b-c))

3). (x/(1-9x^2))*((3x-1)/5x)=(x/(1-3x)(1+3x))*(-(3x+1)/5x^2)=(1/(1-3x))*(-1/5x)=-(1/(5x(1-3x)))=-(1/(5x-15x))

4). ((x^2-9)/(x^2+4x+4))/((3-x)/(3x+6))*(x/(x-3))=(((x-3)*(x+3))/(x^2+4x+4))*((3(x+2))/-(3+x))*(x/(x-3))=-(3x/(x+2))