Question

Доброго времени суток прошу помощи в данных уровнениях,хотя бы до момента сокращения общего знаменателя!30б

Answer 1

$displaystyle tt 1): 5+frac{x+2}{x-4}=frac{48}{x^2-16}: : : : : | : xne4, : xne-4\displaystyle tt frac{x+2}{x-4}-frac{48}{x^2-16}=-5\displaystyle tt frac{x+2}{x-4}-frac{48}{(x-4)(x+4)}=-5\displaystyle tt frac{(x+4)(x+2)-48}{(x-4)(x+4)}=-5\displaystyle tt frac{x^2+2x+4x+8-48}{(x-4)(x+4)}=-5\displaystyle tt frac{x^2+6-40}{(x-4)(x+4)}=-5\displaystyle tt frac{x^2+10x-4x-40}{(x-4)(x+4)}=-5\displaystyle tt frac{x(x+10)-4(x+10)}{(x-4)(x+4)}=-5$

$displaystyle tt frac{(x+10)(x-4)}{(x-4)(x+4)}=-5\displaystyle tt frac{x+10}{x+4}=-5 : : : : | cdot(x+4)\displaystyle tt x+10=-5(x+40)\displaystyle tt x+10=-5x-20\displaystyle tt x+5x=-20-10\displaystyle tt 6x=-30\displaystyle tt x=frac{-30}{6}\displaystyle tt boxed{bold{x=-5}}$

$displaystyle tt 2): frac{x+2}{x-1}-frac{6}{x^2-1}=frac{6}{x+1}: : : : : | : xne1, : xne-1\displaystyle tt frac{x+2}{x-1}-frac{6}{(x-1)(x+1)}-frac{6}{x+1}=0\displaystyle tt frac{(x+1)(x+2)-6-6(x-1)}{(x-1)(x+1)}=0\displaystyle tt frac{x^2+2x+x+2-6-6x+6}{(x-1)(x+1)}=0\displaystyle tt frac{x^2-3x+2}{(x-1)(x+1)}=0\displaystyle tt frac{x^2-x-2x+2}{(x-1)(x+1)}=0\displaystyle tt frac{x(x-1)-2(x-1)}{(x-1)(x+1)}=0\displaystyle tt frac{(x-1)(x-2)}{(x-1)(x+1)}=0$

$displaystyle tt frac{x-2}{x+1}=0\displaystyle tt x-2=0\displaystyle tt boxed{bold{x=2}}$

$displaystyle tt 3): frac{x+3}{x-2}+frac{x-11}{x+2}=frac{20}{x^2-4}: : : : : | : xne2, : xne-2\displaystyle tt frac{x+3}{x-2}+frac{x-11}{x+2}-frac{20}{(x-2)(x+2)}=0\displaystyle tt frac{(x+2)(x+3)+(x-2)(x-11)-20}{(x-2)(x+2)}=0\displaystyle tt frac{x^2+3x+2x+6+x^2-11x-2x+22-20}{(x-2)(x+2)}=0\displaystyle tt frac{2x^2-8x+8}{(x-2)(x+2)}=0\displaystyle tt frac{2(x^2-4x+4)}{(x-2)(x+2)}=0\displaystyle tt frac{2(x-2)^2}{(x-2)(x+2)}=0$

$displaystyle tt frac{2(x-2)}{x+2}=0\displaystyle tt 2(x-2)=0\displaystyle tt 2x-4=0\displaystyle tt 2x=4\displaystyle tt x=frac{4}{2}\displaystyle tt x=2 \displaystyle tt xne2 : to : boxed{bold{xin oslash}}$

$displaystyle tt 4): frac{x-5}{x+1}=frac{24}{(x+1)(x+3)}-frac{9-x}{x-3} : : : : : | : xne-1, : xne3\displaystyle tt frac{x-5}{x+1}-frac{24}{(x+1)(x-3)}+frac{9-x}{x-3}=0\displaystyle tt frac{(x-3)(x-5)-24+(x+1)(9-x)}{(x+1)(x-3)}=0\displaystyle tt frac{x^2-5x-3x+15-24+9x-x^2+9-x}{(x+1)(x-3)}=0\displaystyle tt frac{0+0}{(x+1)(x-3)}=0\displaystyle tt frac{0}{(x+1)(x-3)}=0\displaystyle tt 0=0$