Question

Помогите решить
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Answer 1

$frac{(2x+1)^2}{25} + frac{x-1}{3} =x \ 3(2x+1)^2+25(x-1)=75x \ 3(4x^2+4x+1)+25x-25-75x=0 \ 12x^2+12x+3+25x-25-75x=0 \ 12x^2-38x-22=0 \ 6x^2-19x-11=0 \ D=361+264=625 \ x_1= frac{19+25}{12} = frac{44}{12}= frac{11}{3} \ x_2= frac{19-25}{12} = -frac{6}{12}= -0.5$

$frac{(3x+2)^2}{11} - frac{x+5}{4} =x^2 \ 4(3x+2)^2-11(x+5)=44x^2 \ 4(9x^2+12x+4)-11x-55-44x^2=0 \ 36x^2+48x+16-11x-55-44x^2=0 \ -8x^2+37x-39=0 \ 8x^2-37x+39=0 \ D=1369-1248=121 \ x_1= frac{37+11}{16}=3 \ x_2= frac{37-11}{16}=frac{26}{16}=frac{13}{8}$

$frac{(2-x)^2}{3} - 2x=frac{(7+2x)^2}{5} \ 5(2-x)^2-30x=3(7+2x)^2 \ 5(4-4x+x^2)-30x=3(49+28x+4x^2) \ 20-20x+5x^2-30x-147-84x-12x^2=0 \ -7x^2-134x-127=0 \ 7x^2+134x+127=0 \ D_1=4489-889=3600 \ x_1= frac{-67-60}{7} =-frac{-127}{7} \ x_2= frac{-67+60}{7} =-1$

$frac{(6-x)^2}{8} +x=7- frac{(2x-1)^2}{3} \ 3(6-x)^2+24x=168-8(2x-1)^2 \ 3(36-12x+x^2)+24x=168-8(4x^2-4x+1) \ 108-36x+3x^2+24x-168+32x^2-32x+8 \ 35x^2-44x-52=0 \ D_1=484+2030=2304 \ x_1= frac{22+48}{35} =2 \ x_2= frac{22-48}{35} =- frac{26}{35}$