Question

Помогите решить пожалуйста ​

Answer 1

Решение:

$a) \frac{3x+1}{x+1} + \frac{3+x}{x+1} = \frac{3x+1+3+x}{x+1} = \frac{4x+4}{x+1} = \frac{4(x+1)}{x+1} = 4$

$b) \frac{3x}{x-4} + \frac{x+8}{4-x} = \frac{3x}{x-4} - \frac{x+8}{x-4} = \frac{3x-x-8}{x-4} = \frac{2x-8}{x-4} = \frac{2(x-4)}{x-4} = 2$

$c) \frac{5x+1}{x-1} - \frac{6}{x-1} = \frac{5x+1-6}{x-1} = \frac{5x-5}{x-1} = \frac{5(x-1)}{x-1} = 5$

$d) \frac{5x-6}{x^2-3xy} - \frac{5y-2}{xy-3y^2} = \frac{5x-6}{x(x-3y)} - \frac{5y-2}{y(x-3y)} = \frac{5xy-6y-5xy+2x}{xy(x-3y)} = \frac{2x-6y}{xy(x-3y)} =\frac{2(x-3y)}{xy(x-3y)} =\\\frac{2}{xy}$

Answer 2

а)...=(3х+1+3+х)/(х+1)=4*(х+1)/(х+1)=4

б)... =(3х-х-8)/(х-4)=2*(х-4)/(х-4)=2

в) ...=(5х+1-6)/(х-1)=5*(х-1)/(х-1)=5

д) ...=(5х-6)/х*(х-3у)-(5у-2)/(у*(х-3у))=(5ху-6у-5ху+2х)/(ху*(х-3у))=

2*(х-3у)/(ху*(х-3у))=2/(ху)