Question

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

Answer 1

$\displaystyle \tt 1)\: 5+\frac{x+2}{x-4}=\frac{48}{x^2-16}\: \: \: \: \: | \: x\ne4, \: x\ne-4\\\displaystyle \tt \frac{x+2}{x-4}-\frac{48}{x^2-16}=-5\\\displaystyle \tt \frac{x+2}{x-4}-\frac{48}{(x-4)(x+4)}=-5\\\displaystyle \tt \frac{(x+4)(x+2)-48}{(x-4)(x+4)}=-5\\\displaystyle \tt \frac{x^2+2x+4x+8-48}{(x-4)(x+4)}=-5\\\displaystyle \tt \frac{x^2+6-40}{(x-4)(x+4)}=-5\\\displaystyle \tt \frac{x^2+10x-4x-40}{(x-4)(x+4)}=-5\\\displaystyle \tt \frac{x(x+10)-4(x+10)}{(x-4)(x+4)}=-5\\$

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

$\displaystyle \tt 2)\: \frac{x+2}{x-1}-\frac{6}{x^2-1}=\frac{6}{x+1}\: \: \: \: \: | \: x\ne1, \: x\ne-1\\\displaystyle \tt \frac{x+2}{x-1}-\frac{6}{(x-1)(x+1)}-\frac{6}{x+1}=0\\\displaystyle \tt \frac{(x+1)(x+2)-6-6(x-1)}{(x-1)(x+1)}=0\\\displaystyle \tt \frac{x^2+2x+x+2-6-6x+6}{(x-1)(x+1)}=0\\\displaystyle \tt \frac{x^2-3x+2}{(x-1)(x+1)}=0\\\displaystyle \tt \frac{x^2-x-2x+2}{(x-1)(x+1)}=0\\\displaystyle \tt \frac{x(x-1)-2(x-1)}{(x-1)(x+1)}=0\\\displaystyle \tt \frac{(x-1)(x-2)}{(x-1)(x+1)}=0\\$

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

$\displaystyle \tt 3)\: \frac{x+3}{x-2}+\frac{x-11}{x+2}=\frac{20}{x^2-4}\: \: \: \: \: | \: x\ne2, \: x\ne-2\\\displaystyle \tt \frac{x+3}{x-2}+\frac{x-11}{x+2}-\frac{20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{(x+2)(x+3)+(x-2)(x-11)-20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{x^2+3x+2x+6+x^2-11x-2x+22-20}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{2x^2-8x+8}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{2(x^2-4x+4)}{(x-2)(x+2)}=0\\\displaystyle \tt \frac{2(x-2)^2}{(x-2)(x+2)}=0$

$\displaystyle \tt \frac{2(x-2)}{x+2}=0\\\displaystyle \tt 2(x-2)=0\\\displaystyle \tt 2x-4=0\\\displaystyle \tt 2x=4\\\displaystyle \tt x=\frac{4}{2}\\\displaystyle \tt x=2 \\\displaystyle \tt x\ne2 \: \to \: \boxed{\bold{x\in \oslash}}$

$\displaystyle \tt 4)\: \frac{x-5}{x+1}=\frac{24}{(x+1)(x+3)}-\frac{9-x}{x-3} \: \: \: \: \: | \: x\ne-1, \: x\ne3\\\displaystyle \tt \frac{x-5}{x+1}-\frac{24}{(x+1)(x-3)}+\frac{9-x}{x-3}=0\\\displaystyle \tt \frac{(x-3)(x-5)-24+(x+1)(9-x)}{(x+1)(x-3)}=0\\\displaystyle \tt \frac{x^2-5x-3x+15-24+9x-x^2+9-x}{(x+1)(x-3)}=0\\\displaystyle \tt \frac{0+0}{(x+1)(x-3)}=0\\\displaystyle \tt \frac{0}{(x+1)(x-3)}=0\\\displaystyle \tt 0=0\\$