Question

1. выражение: а) х б) m+2 1 в) х y х - ; - ; + х+1 4m m+4 х+y х-y г) 3х+y х+3y д) х-9 3 - ; - хх(во 2 степени) + xy y(во 2 ст.) +хy х(во 2 3х-х(во 2ст) е) 1 6х ж) 3 21 4-х - ; __ + - х-2 х(в 3 ст)-8 х х(в 2 ст)-7х х-7

Answer 1

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

$\frac{m+2}{4m}-\frac{1}{m+4}=\frac{(m+2)(m+4)-4m}{4m(m+4)}=\frac{m^2+6m+8-4m}{4m(m+4)}=\frac{m^2+2m+8}{4m(m+4)}$

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

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

$\frac{x-9}{x^2-9}-\frac{3}{3x-x^2}=\frac{x-9}{x^2-9}+\frac{3}{x^2-3x}=\frac{x(x-9)+3(x+3)}{x(x-3)(x+3)}=\frac{x^2-9x+3x+9}{x(x-3)(x+3)}= \\ =\frac{x^2-6x+9}{x(x-3)(x+3)}=\frac{(x-3)^2}{x(x-3)(x+3)}=\frac{x-3}{x(x+3)}$

$\frac{1}{x-2}-\frac{6x}{x^3-8}=\frac{x^2+2x+4-6x}{(x-2)(x^2+2x+4)}=\frac{x^2-4x+4}{(x-2)(x^2+2x+4)}=\frac{(x-2)^2}{(x-2)(x^2+2x+4)}= \\ =\frac{x-2}{x^2+2x+4}$

$\frac{3}{x}+\frac{21}{x^2-7x}-\frac{4-x}{x-7}=\frac{3(x-7)+21-x(4-x)}{x(x-7)}=\frac{3x-21+21-4x+x^2}{x(x-7)}= \\ =\frac{x^2-x}{x(x-7)}=\frac{x(x-1)}{x(x-7)}=\frac{x-1}{x-7}$