Question

Алгебра 8 класс,выполните действия

Answer 1

$displaystyle 1)..frac{3-2x}{x^{2}-25}- frac{8-x}{x^{2}-25}= frac{3-2x-8+x}{(x-5)(x+5)}=- frac{5+x}{(x-5)(x+5)}=frac{1}{5-x}; \ \ 2).. frac{y^{2}+30}{y-6}- frac{6-12y}{6-y}= frac{y^{2}+30+6-12y}{y-6}= frac{y^{2}-12y+36}{y-6}= \ \ frac{(y-6)^{2}}{y-6}=y-6;\\3)..frac{xy-2}{xy}- frac{xy}{2+xy}= frac{(xy-2)(xy+2)-x^{2}y^{2}}{xy(2+xy)}=\\= frac{x^{2}y^{2}-2xy+2xy-4-x^{2}y^{2}}{xy(2+xy)}=-frac{4}{xy(2+xy)};$

$displaystyle 4)..frac{1}{1-2x+x^{2}}-frac{1}{x^{2}-x}=frac{1}{(x-1)^{2}}-frac{1}{x(x-1)}=frac{x-x+1}{x(x-1)^{2}}=\\=frac{1}{x(x-1)^{2}};\\5)..frac{9b^{2}+1}{3b-1}-3b+1=frac{9b^{2}+1-(3b+1)(3b-1)}{3b-1}=\\frac{9b^{2}+1-(9b^{2}-1)}{3b-1}=frac{2}{3b-1};$

$displaystyle 6)..frac{1}{(y-x)(y+2)}+frac{1}{(x+2)(x-y)}+ frac{1}{(x+2)(y+2)}=\\=frac{1}{(y-x)(y+2)}-frac{1}{(x+2)(y-x)}+frac{1}{(x+2)(y+2)}=\\=frac{(x+2)-(y+2)+y-x}{(y-x)(x+2)(y+2)}= frac{x+2-y-2+y-x}{(y-x)(x+2)(y+2)}= \ \ = frac{0}{(y-x)(x+2)(y+2)}=0$