Question

помогите решить пожалуйста

Answer 1

$1. \ a) frac{6-x}{x+3}= frac{7x+9}{3+x} \ 6-x=7x+9 \ -8x=3 \ x= -frac{3}{8} \ \ b) frac{8x-3}{5-x} = frac{2x-1}{x-5} \ frac{8x-3}{x-5}=- frac{2x-1}{x-5} \ 8x-3=-2x+1 \ 10x=4 \ x= frac{4}{10}= frac{2}{5} \ \ 2. \ a) frac{19-22x}{13-2x}=-5 \ 19-22x=-5(13-2x) \ 19-22x=-65+10x \ -32x=-84 \ x= frac{84}{32}= frac{21}{8}$

$b) frac{24x^2+37}{2-x}=17 \ 24x^2+37=17(2-x) \ 24x^2+17x+3=0 \ D=289-288=1 \ x_{1} = frac{-17-1}{48}= -frac{3}{8} \ x_{2}= frac{-17+1}{48}=- frac{1}{3} \ \ 3. \ a) frac{x}{1,6-x}= frac{5}{5x+1} \ 5x(5x+1)=5(8-5x) \ 5(5x^2+6x-8)=0 \ 5x^2+6x-8=0 \ D=36+160=196=14^2 \ x_{1}= frac{-6-14}{10}=-2 \ x_{2}= frac{-6+14}{10}= frac{8}{10}= frac{4}{5}$

$b) frac{3x+4}{1-3x^2}= frac{4}{x} \ (3x+4)x=4(1-3x^2) \ 15x^2+4x-4=0 \ D=16+240=256=16^2 \ x_{1} = frac{-4-16}{30}=- frac{2}{3} \ x_{2}= frac{-4+16}{30}= frac{2}{5} \ \ 4. \ frac{3-3x}{x}=11-11x \ 3-3x=x(11-11x)} \ 11x^2-14x+3=0 \ D=196-132=64=8^2 \ x_{1}= frac{14-8}{22} = frac{3}{11} \ x_{2}= frac{14+8}{22}=1$