Людиии пж помогите. прошуу номер 343(1.3.5)

Решите задачу:

$1) \frac{(x+1)(x-1)+120}{6(x-1)} = \frac{4*6(x-1)}{6(x-1)} \\ x^{2} -1+120=24x-24 \\ x^{2} -24x+143=0 \\ D=576-4*143=4\ \textgreater \ 0 \\ x1= \frac{24+2}{2} =13 \\ x2= \frac{24-2}{2} =11$
$3) \frac{3(x+1)+(1-x)}{(1-x)(x+1)} = \frac{28}{(1-x)(1+x)} \\ 3x+3+1-x=28 \\ 2x=24 \\ x=12$
$5) \frac{16(1-x)+30(x-3)}{(x-3)(1-x)}= \frac{3*(x-3)(1-x)}{(x-3)(1-x)} \\ 16-16x+30x-90=3x-3x^2-9+9x \\ 3x^2+2x-65=0 \\ D=4-4*3*(-65)=784\ \textgreater \ 0 \\ x1= \frac{-2+28}{6}= \frac{26}{6}= 4\frac{1}{3} \\ x2= \frac{-2-28}{6} =-5$