Question

Помогите пожалуйста разобраться

Answer 1

$20)\ \ 1.\ \dfrac{3a}{10}+\dfrac{2a}{10}=\dfrac{3a+2a}{10}=\dfrac{5a}{10}\\\\2.\ \dfrac{6x}{5y}-\dfrac{x}{5y}=\dfrac{6x-x}{5y}=\dfrac{5x}{5y}=\dfrac{x}{y}\\\\3.\ \dfrac{2m-4n}{21c}+\dfrac{5m+18n}{21c}=\dfrac{(2m-4n)+(5m+18n)}{21c}=\dfrac{7m+14n}{21c}=\\\\=\dfrac{7\, (m+2n)}{21c}=\dfrac{m+2n}{3c}\\\\4.\ \dfrac{2a+5b}{ab}-\dfrac{2a-3b}{ab}=\dfrac{2a+5b-2a+3b}{ab}=\dfrac{8b}{ab}=\dfrac{8}{a}$

$5.\ \dfrac{5y}{y^2-9}-\dfrac{15}{y^2-9}=\dfrac{5y-15}{(y-3)(y+3)}=\dfrac{5\, (y-5)}{(y-5)(y+5)}=\dfrac{5}{y+5}\\\\6.\ \dfrac{y^2+8y}{4-y^2}-\dfrac{4y-4}{4-y^2}=\dfrac{y^2+8y-4y+4}{4-y^2}=\dfrac{y^2-4y+4}{(2-y)(2+y)}=\dfrac{(2-y)^2}{(2-y)(2+y)}=\\\\=\dfrac{2-y}{2+y}$

$21)\ \ 1.\ \dfrac{x-4}{x-2}-\dfrac{x}{2-x}=\dfrac{x-4}{x-2}+\dfrac{x}{x-2}=\dfrac{x-4+x}{x-2}=\dfrac{2x-4}{x-2} =\dfrac{2\, (x-2)}{x-2}=2\\\\2.\ \dfrac{5x+6}{5-x}+\dfrac{3x+16}{x-5}=\dfrac{5x+6}{5-x}-\dfrac{3x+16}{5-x}=\dfrac{5x+6-3x-16}{5-x}=\dfrac{2x-10}{5-x}=\\\\=\dfrac{2\, (x-5)}{-(x-5)}=-2\\\\3.\ \dfrac{(2a-1)^2}{6a-6}+\dfrac{(a-2)^2}{6-6a}=\dfrac{4a^2-4a+1}{6a-6}-\dfrac{a^2-4a+4}{6a-6}=\\\\=\dfrac{4a^2-4a+1-a^2+4a-4}{6a-6}=\dfrac{3a^2-3}{6(a-1)}=\dfrac{3(a-1)(a+1)}{6\, (a-1)}=\dfrac{3(a+1)}{6}$

$4.\ \dfrac{16-7x}{(x-4)^2}-\dfrac{x-x^2}{(4-x)^2}=\dfrac{16-7x-x+x^2}{(x-4)^2}=\dfrac{x^2-8x+16}{(x-4)^2}=\\\\=\dfrac{(x-4)^2}{(x-4)^2}=1\ \ ,\ \ \ \ (x-4)^2=(4-x)^2\ \ (!)\\\\\\22)\ \ \dfrac{y+4}{y}=\dfrac{y}{y}+\dfrac{4}{y}=1+\dfrac{4}{y}\\\\\dfrac{a^2-3a+4}{a-3}=\dfrac{a(a-3)}{a-3}+\dfrac{4}{a-3}=a+\dfrac{4}{a-3}\\\\\dfrac{x^2+4x-8}{x-4}=\dfrac{(x^2-16)+(4x+8)}{x-4}=\dfrac{(x-4)(x+4)+(4x+8)}{x-4}=\\\\=\dfrac{(x-4)(x+4)}{x-4}+\dfrac{4(x+2)}{x-4}=(x+4)+\dfrac{4(x+2)}{x-4}$