Question

упростите выражения
1) (x+3/x)-(x/x-3)+(9/x²-3x)
2) (b+c/b²-bc)-(4b/b²-c²)-(b-c/b²+bc)
3) (1/x-2)-(6x/x³-8)+(x-2/x²+2x+4)

Answer 1

Если $x neq 0,and,x neq 3$, то:
$frac{x+3}{x}- frac{x}{x-3}+ frac{9}{x^2-3x}= frac{(x+3)*(x-3)}{x*(x-3)}- frac{x*x}{(x-3)*x}+ frac{9}{x(x-3)}=$
$= frac{(x+3)*(x-3)-x*x+9}{x(x-3)}= frac{x^2-3^2-x^2+9}{x(x-3)} = frac{0}{x(x-3)} =0$

$frac{b+c}{b^2-bc}- frac{4b}{b^2-c^2}- frac{b-c}{b^2+bc}= frac{b+c}{b(b-c)}- frac{4}{(b-c)(b+c)}- frac{b-c}{b(b+c)}=$
$= frac{(b+c)*(b+c)}{b(b-c)(b+c)}- frac{4b*b}{b*(b-c)(b+c)}- frac{(b-c)*(b-c)}{b(b+c)*(b-c)}=$
$= frac{(b+c)^2}{b(b-c)(b+c)}- frac{4b^2}{b(b-c)(b+c)}- frac{(b-c)^2}{b(b+c)(b-c)}=$
$= frac{(b+c)^2-4b^2-(b-c)^2}{b(b-c)(b+c)} = frac{b^2+2bc+c^2-4b^2-(b^2-2bc+c^2)}{b(b-c)(b+c)}=$
$= frac{b^2+2bc+c^2-4b^2-b^2+2bc-c^2}{b(b-c)(b+c)} = frac{4bc-4b^2}{b(b-c)(b+c)}= frac{-4b(b-c)}{b(b-c)(b+c)}=-frac{4}{b+c}$

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