Question

Помогите решить 1123(а,в)

Answer 1

решаем 1123
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Answer 2

$frac{12x^2+1}{2x+3}=4x-1 \ 12x^2+1=(4x-1)(2x+3) \ 12x^2+1=8x^2+12x-2x-3 \ 12x^2+1=8x^2+10x-3 \ 12x^2+1-8x^2-10x+3=0 \ 4x^2-10x+4=0 |:4 \ x^2-2,5x+1=0 \ D=b^2-4ac \ D=2,5^2-4*1*1=6,25-4=2,25 \ x_{1,2} = frac{2,5+/- sqrt{2,25} }{2} \ x_1 = frac{2,5+1,5}{2} = frac{4}{2} = 2 \ x_2 = frac{2,5-1,5}{2} = frac{1}{2} = 0,5 \ \ 3y+ frac{2y^2+y}{4y+10}=-3 \ frac{3y(4y+10)+(2y^2+y)}{4y+10} = -3 \ frac{12y^2+30y+2y^2+y}{4y+10}=-3 \ frac{14y^2+31y)}{4y+10}=-3$
$14y^2+31y=-3(4y+10) \ 14y^2+31y=-12y-30 \ 14y^2+31y+12y+30=0 \ 14y^2+43y+30=0 \ D=43^2-4*30*14=169 \ x_{1,2} frac{-b+/- sqrt{D} }{2a} \ x_{1,2} frac{-43+/-13}{28} \ x_1 = frac{-43+13}{28} = frac{-30}{28} \ x_2 = frac{-43-13}{28} = frac{-56}{28} = -2 \ \ x_1 = -1frac{1}{14} ; x_2 = -2$