Question

Помогите пожалуйста, алгебра! ​

Answer 1

1) $(2y - 3) * (5y +1) = 2y +\frac{2}{5}$

$10y^{2} + 2y -15y -3 = 2y +\frac{2}{5}$

$10y^{2} + 2y -15y -3 - 2y -\frac{2}{5}=0$

$10y^{2} -15y -3\frac{2}{5}=0$

$10y^{2} -15y -\frac{17}{5}=0$

$10y^{2}*5 -15y*5 -\frac{17}{5}*5=0$

$50y^{2} -75y - 17 =0$

$50y^{2} + 10y -85y - 17 =0$

$10y(5y+1) - 17 (5y +1) =0$

$(5y +1)*(10y-17) =0$

$5y + 1 =0$                                       $10y-17 =0$

$y = \frac{-1}{5}= - \frac{1}{5}$           $y =\frac{17}{10}=1\frac{7}{10}$  

2) $-y*(y+7) = (2+y)*(y-2)$

$-y^2-7y = 2y-4+y^2-2y$

$-y^2-7y - 2y+4-y^2+2y=0$

$-2y^2-7y+4=0$

$2y^2+7y-4=0$

$2y^2+8y-y-4=0$

$2y(y+4)-1(y+4)=0$

$(y+4)*(2y-1)=0$

$y + 4 =0$                               $2y-1 =0$

$y = -4$                                  $y =\frac{1}{2}$  

3) $(3y-1)* (y+3) = y*(6y+1)$

$3y^2 + 9y -y -3= 6y^2+y$

$3y^2 + 9y -y -3 - 6y^2- y =0$

$-3y^2 + 7y -3  =0$

$3y^2 - 7y +3  =0$

$D = (-7)^2 - 4*3*3= 49 - 36 = 13$

$y_{1} = \frac{7+\sqrt{13}}{2*3} = \frac{7+\sqrt{13}}{6}; \\y_{2} = \frac{7-\sqrt{13}}{2*3}  = \frac{7-\sqrt{13}}{6}$

4) $(y+1) * (y-1) = 2(5y+10,5)$

$y^2 - 1 = 10y+21$

$y^2 - 1 - 10y-21=0$

$y^2 - 10y-22=0$

$D = (-10)^2 - 4*1*(-22)= 100 +88 = 188$

$y_{1} = \frac{10+\sqrt{188}}{2*1} = \frac{10+2\sqrt{47}}{2} = 5 +\sqrt{47} ; \\y_{2} = \frac{10-\sqrt{188}}{2*1} = \frac{10-2\sqrt{47}}{2} = 5 -\sqrt{47}$