Question

помогите пожалуйста решить :упростите выражение: а) 1/x2y-xy2 - 3/x3-y3 б)10/x2-10x+25 + 10/x2-25 + 1/x+5 в)5x-1/x2-1 + 2/1-x -

3x/x+1 /- дробь 2,3- степень 1,3,10,5,2- коэффициенты. ЗАРАНЕЕ СПАСИБО!!!

Answer 1

а) $\frac{1}{x^2y-xy^2} - \frac{3}{x^3-y^3}=\frac{1}{xy(x-y)} - \frac{3}{(x-y)(x^2+xy+y^2)}=\\\ =\frac{x^2+xy+y^2-3xy}{xy(x-y)(x^2+xy+y^2)}=\frac{x^2-2xy+y^2}{xy(x-y)(x^2+xy+y^2)}=\\\ =\frac{(x-y)^2}{xy(x-y)(x^2+xy+y^2)}=\frac{x-y}{xy(x^2+xy+y^2)}$

б) $\frac{10}{x^2-10x+25} + \frac{10}{x^2-25} + \frac{1}{x+5}=\frac{10}{(x-5)^2} + \frac{10}{(x-5)(x+5)} + \frac{1}{x+5}=\\\ = \frac{10(x+5)+10(x-5)+(x-5)^2}{(x+5)(x-5)^2}=\frac{10x+50+10x-50+x^2-10x+25}{(x+5)(x-5)^2}=\\\ =\frac{x^2+10x+25}{(x+5)(x-5)^2}=\frac{(x+5)^2}{(x+5)(x-5)^2}=\frac{x+5}{(x-5)^2}$

в) $\frac{5x-1}{x^2-1}+\frac{2}{1-x}-\frac{3x}{x+1}=\frac{5x-1}{(x-1)(x+1)}-\frac{2}{x-1}-\frac{3x}{x+1}=\\\ =\frac{5x-1-2(x+1)-3x(x-1)}{(x-1)(x+1)}=\frac{5x-1-2x-2-3x^2+3x}{(x-1)(x+1)}=\\\ =\frac{-3x^2+6x-3}{(x-1)(x+1)}=-\frac{3(x^2-2x+1)}{(x-1)(x+1)}=-\frac{3(x-1)^2}{(x-1)(x+1)}=\\\ =-\frac{3(x-1)}{x+1}=$