Question

4 раз викладаю допоможіть

Answer 1

$1)\\\\\frac{2a-5b}{4ab}+\frac{13b-2a}{4ab}=\frac{2a-5b+13b-2a}{4ab}=\frac{8b}{4ab}=\frac{2}{a}\\\\2)\\\\\frac{7m+y^5}{5y}-\frac{7m-4y^5}{5y}=\frac{7m+y^5-(7m-4y^5)}{5y}=\frac{7m+y^5-7m+4y^5}{5y}=\frac{5y^5}{5y}=y^4\\\\3)\\\\\frac{7x-t^4}{8t^5}-\frac{t^4+7x}{8t^5}=\frac{7x-t^4-(t^4+7x)}{8t^5}=\frac{7x-t^4-t^4-7x}{8t^5}=\frac{-2t^4}{8t^5}=-\frac{1}{4t}$

$4)\\\\\frac{x-3}{4x}+\frac{2x+5}{4x}-\frac{2-x}{4x}=\frac{x-3+2x+5-(2-x)}{4x}=\frac{x-3+2x+5-2+x}{4x}=\frac{4x}{4x}=1\\\\5)\\\\\frac{3a-7}{12a}-\frac{5a-13}{12a}+\frac{2-6a}{12a}=\frac{3a-7-(5a-13)+2-6a}{12a}=\frac{3a-7-5a+13+2-6a}{12a}=\\\\\frac{8 - 8a}{12a}=\frac{4(2-2a)}{12a}=\frac{2-2a}{3a}\\\\6)\\\\\frac{3y+8}{4y}+\frac{2y-11}{4y}-\frac{y-3}{4y}=\frac{3y+8+2y-11-(y-3)}{4y}=\frac{3y+8+2y-11-y+3}{4y}=\\\\\frac{3y+8+2y-11-y+3}{4y}=\frac{4y}{4y}=1$