Здравствуйте! Помогите пожалуйста привести дробь к общему знаменателю: а)4/2х+1 - 2х+5/2х+1 б)6х+7/1-2х + 5х+4/2х-1 в)3/х + 4х-3/х-1 г)3х-1/х²+2х - 3/х+2 + 5/х д)2/х²-9 + 1/х-3 + 2/х+3 е)2/х+1 - х+3/х+1 ё)7х+5/1-х + 3х+6/х-1 ж)...

а) [latex] \frac{4}{2x+1} - \frac{2x+5}{2x+1} = \frac{4-2x+5}{2x+1} = \frac{9-2x}{2x+1} [/latex] б) [latex] \frac{6x+7}{1-2x} + \frac{5x+4}{2x-1} = \frac{6x+7}{1-2x} - \frac{5x+4}{1-2x} = \frac{6x+7-5x-3}{1-2x} = \frac{x+4}{1-2x} [/latex] в) [latex] \frac{3}{x} + \frac{4x-3}{x-1} = \frac{3(x-1)}{x(x-1)} + \frac{x(4x-3)}{x(x-1)} = \frac{3x-3}{x^2-x} + \frac{4x^2-3x}{x^2-x} = \frac{3x-3+4x^2-3x}{x^2-x} = \frac{4x^2-3}{x^2-x} [/latex] г) [latex] \frac{3x-1}{x^2+2x} - \frac{3}{x+2} + \frac{5}{x} = \frac{3x-1}{x^2+2x} - \frac{3x}{x^2+2x} + \frac{5(x+2)}{x^2+2x} = \frac{3x-1 - 3x+ 5x+10}{x^2+2x} = \frac{5x+9}{x^2+2x} [/latex] д) [latex] \frac{2}{x^2-9} + \frac{1}{x-3} + \frac{2}{x+3} = \frac{2}{x^2-9} + \frac{x+3}{x^2-9} + \frac{2(x-3)}{x^2-9} = \frac{2+x+3+2x-6}{x^2-9} = \frac{3x-1}{x^2-9} [/latex] е) [latex] \frac{2}{x+1} - \frac{x+3}{x+1} = \frac{2-x-3}{x+1} = \frac{-x-1}{x+1} = - \frac{x+1}{x+1} = -1 [/latex] ё) [latex] \frac{7x+5}{x-1} + \frac{3x+6}{1-x} = \frac{7x+5}{x-1} - \frac{3x+6}{x-1} = \frac{7x+5-3x-6}{x-1} = \frac{4x-1}{x-1} [/latex] ж) [latex] \frac{2}{x} + \frac{3x-2}{x+1} = \frac{2(x+1)}{x^2+x} + \frac{x(3x-2)}{x^2+x} = \frac{2x+2+3x^2-2x}{x^2+x} = \frac{3x^2+2}{x^2+x} [/latex] з) [latex] \frac{x+3}{x^2+x} - \frac{1}{x+1} + \frac{2}{x} = \frac{x+3}{x^2+x} - \frac{x}{x^2+x} + \frac{2(x+1)}{x^2+x} = \frac{x+3-x+2x+2}{x^2+x} = \frac{2x+5}{x^2+x} [/latex] и) [latex] \frac{4}{x^2-4} - \frac{1}{x-2} - \frac{1}{x+2} = \frac{x}{x^2-4} - \frac{x+2}{x^2-4} - \frac{x-2}{x^2-4} = \frac{x-x-2-x+2}{x^2-4} = \frac{-x}{x^2-4} [/latex] Удачи!