Question

НА СТО БАЛЛОВ
Решите только тот номер что обведен

Answer 1

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

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

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

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

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

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