Question

Помогите пожалуйста! ​

Answer 1

35.

$1) (\frac{a-2}{a+2}-\frac{a+2}{a-2} ):\frac{12a^2}{4-a^2}= (\frac{(a-2)*(a-2)}{(a+2)(a-2)}-\frac{(a+2)(a+2)}{(a-2)(a+2)} ):\frac{12a^2}{4-a^2}=\frac{(a-2)*(a-2)-(a+2)(a+2)}{(a+2)(a-2)}:\frac{12a^2}{4-a^2}=\frac{a^2-4a+4-(a^2+4a+4)}{(a+2)(a-2)}:\frac{12a^2}{4-a^2}=\frac{a^2-4a+4-a^2-4a-4}{(a+2)(a-2)}*\frac{4-a^2}{12a^2}=\frac{-8a}{(a+2)(a-2)}*\frac{(2-a)(2+a)}{12a^2}=\frac{-8a(2-a)(2+a)}{(a+2)(a-2)*12a^2}=\frac{8a(a-2)(2+a)}{(a+2)(a-2)*12a^2}=\frac{2}{3a}$

$2)(\frac{8x}{x-2}+2x ):\frac{4x+8}{7x-14}= (\frac{8x}{x-2}+\frac{2x*(x-2)}{x-2} ):\frac{4x+8}{7x-14}=\frac{8x+2x(x-2)}{x-2}:\frac{4x+8}{7x-14}=\frac{8x+2x^2-4x}{x-2}:\frac{4(x+2)}{7(x-2)}=\frac{2x^2+4x}{x-2}*\frac{7(x-2)}{4(x+2)}=\frac{2x(x+2)*7(x-2)}{(x-2)*4(x+2)}=\frac{7x}{2}$

$3) \frac{5a}{a+3} +\frac{a-6}{3a+9}*\frac{135}{6a-a^2} = \frac{5a}{a+3} +\frac{(a-6)*135}{3(a+3)*a(6-a)}=\frac{5a}{a+3} +\frac{-45}{a(a+3)}=\frac{5a*a}{a(a+3)} +\frac{-45}{a(a+3)}=\frac{5a^2+-45}{a(a+3)} =\frac{5(a^2-9)}{a(a+3)} =\frac{5(a-3)(a+3)}{a(a+3)} =\frac{5(a-3)}{a}$

$4) (\frac{3m}{m+5} -\frac{8m}{m^2+10m+25} ):\frac{3m+7}{m^2-25}+\frac{5m-25}{m+5}\\1. \frac{3m}{m+5} -\frac{8m}{m^2+10m+25} =\frac{3m}{m+5} -\frac{8m}{(m+5)^2} =\frac{3m*(m+5)}{(m+5)*(m+5)} -\frac{8m}{(m+5)^2}=\frac{3m*(m+5)-8m}{(m+5)^2}=\frac{3m^2+15m-8m}{(m+5)^2}=\frac{3m^2+7m}{(m+5)^2}=\frac{m(3m+7)}{(m+5)^2}\\2.\frac{m(3m+7)}{(m+5)^2}:\frac{3m+7}{m^2-25}=\frac{m(3m+7)}{(m+5)^2}:\frac{3m+7}{(m-5)(m+5)}=\frac{m(3m+7)}{(m+5)^2}*\frac{(m-5)(m+5)}{3m+7}=\frac{m(m-5)}{m+5}$

$3. \frac{m(m-5)}{m+5}+\frac{5m-25}{m+5} =\frac{m(m-5)+5(m-5)}{m+5} =\frac{(m-5)(m+5)}{m+5}=m-5$

$5) (\frac{y^2}{x^3-xy^2}+\frac{1}{x+y} ):(\frac{x-y}{x^2+xy} -\frac{x}{xy+y^2} )\\1. \frac{y^2}{x^3-xy^2}+\frac{1}{x+y}=\frac{y^2}{x(x^2-y^2)}+\frac{1}{x+y}=\frac{y^2}{x(x-y)(x+y)}+\frac{1}{x+y}=\frac{y^2}{x(x-y)(x+y)}+\frac{1*x(x-y)}{(x+y)*x(x-y)}=\frac{y^2+x^2-xy}{x(x-y)(x+y)}\\2. \frac{x-y}{x^2+xy} -\frac{x}{xy+y^2}=\frac{x-y}{x(x+y)} -\frac{x}{y(x+y)}=\frac{(x-y)*y}{x(x+y)*y} -\frac{x*x}{y(x+y)*x}=\frac{(x-y)*y-x*x}{xy(x+y)}=\frac{xy-y^2-x^2}{xy(x+y)}=\frac{-(y^2+x^2-xy)}{xy(x+y)}$$3.\frac{y^2+x^2-xy}{x(x-y)(x+y)}:\frac{-(y^2+x^2-xy)}{xy(x+y)} =\frac{y^2+x^2-xy}{x(x-y)(x+y)}*\frac{xy(x+y)}{-(y^2+x^2-xy)}=-\frac{y}{x-y} =\frac{y}{y-x}$

$6) (\frac{a}{a-4} -\frac{a}{a+4}-\frac{a^2+16}{16-a^2} ):\frac{4a+a^2}{(4-a)^2}\\1. \frac{a}{a-4} -\frac{a}{a+4}=\frac{a*(a+4)}{(a-4)*(a+4)} -\frac{a*(a-4)}{(a+4)*(a-4)}=\frac{a*(a+4)-a(a-4)}{(a-4)*(a+4)} =\frac{a^2+4a-a^2+4a}{(a-4)*(a+4)}=\frac{8a}{(a-4)(a+4)}$

$2.\frac{8a}{(a-4)(a+4)}-\frac{a^2+16}{16-a^2} =\frac{8a}{(a-4)(a+4)}-\frac{a^2+16}{(4-a)(4+a)}=\frac{8a}{(a-4)(a+4)}-\frac{a^2+16}{-(a-4)(4+a)}=\frac{8a}{(a-4)(a+4)}+\frac{a^2+16}{(a-4)(4+a)}=\frac{8a+a^2+16}{(a-4)(a+4)}=\frac{(a+4)^2}{(a-4)(a+4)}=\frac{a+4}{(a-4)}$

$3. \frac{a+4}{(a-4)}:\frac{4a+a^2}{(4-a)^2}= \frac{a+4}{(a-4)}:\frac{a(4+a)}{(4-a)^2} =\frac{a+4}{(a-4)}*\frac{(4-a)^2}{a(4+a)} =\frac{a-4}{a}$