Question

ХЭЛП АЛГЕБРА 8 КЛАСС

Answer 1

1)

$1)\\\frac{a}{a-b}+\frac{a}{b}=\frac{ab}{b(a-b)}+\frac{a(a-b)}{b(a-b)}=\frac{ab+a(a-b)}{b(a-b)}=\frac{ab+a^2-ab}{b(a-b)}=\frac{a^2}{b(a-b)}\\\\2)\\\frac{4}{x}-\frac{5x+4}{x+2}=\frac{4(x+2)}{x(x+2)}-\frac{x(5x+4)}{x(x+2)}=\frac{4(x+2)-x(5x+4)}{x(x+2)}=\frac{4x+8-5x^2-4x}{x(x+2)}\\\\=\frac{8-5x^2}{x(x+2)}\\\\3)\\\frac{b}{b-2}-\frac{2}{b+2}=\frac{b(b+2)}{(b-2)(b+2)}-\frac{2(b-2)}{(b-2)(b+2)}=\frac{b(b+2)-2(b-2)}{(b-2)(b+2)}=\\\\\frac{b^2+2b-2b+4}{(b-2)(b+2)}=\frac{b^2+4}{(b-2)(b+2)}$

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2)

$1)\\\frac{b}{b-5}-\frac{4b-1}{4b-20}=\frac{b}{b-5}-\frac{4b-1}{4(b-5)}=\frac{4b}{4(b-5)}-\frac{4b-1}{4(b-5)}=\\\\\frac{4b-(4b-1)}{4(b-5)}=\frac{4b-4b+1}{4(b-5)}=\frac{1}{4(b-5)}$

$2)\\\frac{a-2}{2a-6}-\frac{a-1}{3a-9}=\frac{a-2}{2(a-3)}-\frac{a-1}{3(a-3)}=\\\\\frac{3(a-2)}{6(a-3)}-\frac{2(a-1)}{6(a-3)}=\frac{3(a-2)-2(a-1)}{6(a-3)}=\\\\\frac{3a-6-2a+2}{6(a-3)}=\frac{a-4}{6(a-3)}$

$3)\\\frac{b+4}{ab-b^2}-\frac{a+4}{a^2-ab}=\frac{b+4}{b(a-b)}-\frac{a+4}{a(a-b)}=\\\\\frac{a(b+4)}{ab(a-b)}-\frac{b(a+4)}{ab(a-b)}=\frac{a(b+4)-b(a+4)}{ab(a-b)}=\\\\\frac{ab+4a-ab-4b}{ab(a-b)}=\frac{4a-4b}{ab(a-b)}=\frac{4(a-b)}{ab(a-b)}=\frac{4}{ab}$

$4)\\\frac{4x-y}{x^2-y^2}+\frac{1}{x-y}=\frac{4x-y}{(x-y)(x+y)}+\frac{x+y}{(x-y)(x+y)}=\\\\\frac{4x-y+x+y}{(x-y)(x+y)}=\frac{5x}{(x-y)(x+y)}$

$5)\\\frac{y^2}{y^2-81}-\frac{y}{y+9}=\frac{y^2}{(y+9)(y-9)}-\frac{y(y-9)}{(y+9)(y-9)}=\\\\\frac{y^2-y(y-9)}{(y+9)(y-9)}=\frac{y^2-y^2+9y}{(y+9)(y-9)}=\frac{9y}{(y+9)(y-9)}$

$6)\\\frac{10a}{25a^2-9}-\frac{1}{5a+3}=\frac{10a}{(5a+3)(5a-3)}-\frac{5a-3}{(5a+3)(5a-3)}=\\\\\frac{10a-(5a-3)}{(5a+3)(5a-3)}=\frac{10a-5a+3}{(5a+3)(5a-3)}=\frac{5a+3}{(5a+3)(5a-3)}=\frac{1}{5a-3}$