Question

Помоги пожалуйста!!!!!все в фото!!!

Answer 1

$frac{x^2+12}{x^2-4}- frac{x+6}{x-2}= frac{x^2+12-(x+6)(x+2)}{x^2-4}= frac{x^2+12-x^2-6x-2x-12}{x^2-4}=frac{-8x}{x^2-4}$
$frac{3-n}{nm-n^2} - frac{3-m}{m^2-n^2} = frac{(3-n)(m+n)-(3-m)(m+n)}{n(m^2-n^2)} = \ = frac{3m+3n-mn-n^2-3m+m^2-3n+mn}{n(m^2-n^2)} = frac{1}{n}$
$frac{42}{4a^2-9} + frac{8}{2a+3} - frac{7}{2a-3} = frac{42+8(2a-3)-7(2a+3)}{4a^2-9} \ = frac{42+16a-24-14a-21}{4a^2-9} = frac{2a-3}{4a^2-9} = frac{1}{2a+3}$

Answer 2

б) $frac{x^2+12}{x^2-4} - frac{x+6}{x-2} = frac{x^2+12-(x+6)(x+2)}{(x-2)(x+2)} = frac{x^2+12-x^2-2x-6x-12}{x^2-4} = -frac{8x}{x^2-4}$

г)$frac{3-n}{nm-n^2} - frac{3-m}{m^2-n^2}= frac{3-n}{n(m-n)} - frac{3-m}{(m-n)(m+n)} = frac{(3-n)(m+n)-n(3-m)}{n(m^2-n^2)}$$= frac{3m+3n-mn-n^2-3n+mn}{n(m^2-n^2)} = frac{3m-n^2}{n(m^2-n^2)}$

е) $frac{42}{4a^2-9} + frac{8}{2a+3}- frac{7}{2a-3} = frac{42+16a-24-14a-21}{(2a-3)(2a+3)} = frac{2a-3}{(2a-3)(2a+3)}= frac{1}{2a+3}$