Question

Срочнооооо!!!!! Мне очень надо, даю 27 баллов

Answer 1

Объяснение:

1.  $\frac{x-2}{x+3}-\frac{30}{x^2-9}=\frac{1}{6}$

$\frac{x-2}{x+3}-\frac{30}{(x-3)(x+3)}=\frac{1}{6}$

$\frac{(x-3)(x-2)-30}{(x-3)(x+3)}=\frac{1}{6}$

$\frac{x^2-2x-3x+6-30}{(x-3)(x+3)}=\frac{1}{6}$

$\frac{x^2-5x-24}{(x-3)(x+3)}=\frac{1}{6}$

$\frac{x^2+3x-8x-24}{(x-3)(x+3)}=\frac{1}{6}$

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

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

$\frac{x-8}{x-3}=\frac{1}{6}$

6(x-8)=x-3

6x-48=x-3

6x-x=-3+48

5x=45

x=9

2. $\frac{x-2}{x+1}-\frac{5}{x-1}=\frac{6}{x^2-1}$

$\frac{x-2}{x+1}-\frac{5}{x-1}-\frac{6}{x^2-1}=0$

$\frac{x-2}{x+1}-\frac{5}{x-1}-\frac{6}{(x-1)(x+1)}=0$

$\frac{(x-1)(x-2)+5(x+1)-6}{(x-1)(x+1)}=0$

$\frac{x^2-2x-x+2+5x+5-6}{(x-1)(x+1)}=0$

$\frac{x^2+2x+1}{(x-1)(x+1)}=0$

$\frac{(x+1)^2}{(x-1)(x+1)}=0$

$\frac{x+1}{x-1}=0$

x+1=0

x=-1

3. $\frac{x-9}{x+1}-\frac{x+3}{1-x}=\frac{8}{x^2-1}$

$\frac{x-9}{x+1}-\frac{x+3}{1-x}-\frac{8}{x^2-1}=0$

$\frac{x-9}{x+1}-\frac{x+3}{-(x-1)}-\frac{8}{(x-1)(x+1)}=0$

$\frac{x-9}{x+1}+\frac{x+3}{x-1}-\frac{8}{(x-1)(x+1)}=0$

$\frac{(x-1)(x-9)+(x+1)(x+3)-8}{(x-1)(x+1)}=0$

$\frac{x^2-9x-x+9+x^2+3x+x+3-8}{(x-1)(x+1)}=0$

$\frac{2x^2-6x+4}{(x-1)(x+1)}=0$

$\frac{2(x^2-3x+2)}{(x-1)(x+1)}=0$

$\frac{2(x^2-x-2x+2)}{(x-1)(x+1)}=0$

$\frac{2(x(x-1)-2(x-1))}{(x-1)(x+1)}=0$

$\frac{2(x-1)(x-2)}{(x-1)(x+1)}=0$

$\frac{2(x-2)}{x+1}=0$

2(x-2)=0

x-2=0

x=2

4. $\frac{x+1}{x-2}+\frac{9}{(x-2)(x-5)}=\frac{x-2}{x-5}$

$\frac{x+1}{x-2}+\frac{9}{(x-2)(x-5)}-\frac{x-2}{x-5}=0$

$\frac{(x-5)(x+1)+9-(x-2)^2}{(x-2)(x-5)}=0$

$\frac{x^2+x-5x-5+9-(x^2-4x+4)}{(x-2)(x-5)}=0$

$\frac{x^2-4x+4-(x-2)^2}{(x-2)(x-5)}=0$

$\frac{(x-2)^2-(x-2)^2}{(x-2)(x-5)}=0$

$\frac{(x-2-(x-2))(x-2+(x-2))}{(x-2)(x-5)}=0$

$\frac{(x-2-x+2)(x-2+x-2)}{(x-2)(x-5)}=0$

$\frac{(0+0)(2x-4)}{(x-2)(x-5)}=0$

$\frac{0*2(x-2)}{(x-2)(x-5)}=0$

$\frac{0*2}{x-5}=0$

$\frac{0}{x-5}=0$

0=0