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:
$\frac{x+3}{x}- \frac{x}{x-3}+ \frac{9}{x^2-3x}= \frac{(x+3)*(x-3)}{x*(x-3)}- \frac{x*x}{(x-3)*x}+ \frac{9}{x(x-3)}=$
$= \frac{(x+3)*(x-3)-x*x+9}{x(x-3)}= \frac{x^2-3^2-x^2+9}{x(x-3)} = \frac{0}{x(x-3)} =0$

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

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