Question

Помогите решить уравнения по алгебре! 7 класс

Answer 1

$(2+3y)^2=0 => 2 + 3y=0 => 3y=-2 => y = -\frac{2}{3}\\ (4-2b)^2=0 => 4 - 2b = 0 => 2b = 4 => b = 2\\64-x^2 = 0 => x^2 = 64 => |x| = 8 => x = \pm8\\2(x-2)+x(x-2)=0 => (x-2)(x+2) = 0 => x^2 - 4 = 0 => x^2 = 4 => |x| = 2 => x = \pm2\\4x^2-4x+1=0=>(2x-1)^2=0=>2x-1=0=>2x=1=>x=\frac{1}{2}\\ 2(5+x)+7x(5+x)=0=>(x+5)(7x+2)=0 => x_1=-5; x_2=-\frac{2}{7}\\ 12x^2-2x=0=>2x(6x-1)=0 => x_1=0; x_2=\frac{1}{6}\\15x+3x^2=0 => 3x(5+x)=0 => x_1=0; x_2 = -5\\ z(2-z)(3-2z) =0 => z_1 = 0; z_2 = 2; z_3 = \frac{3}{2}$