Question

Найдите корни уравнений :1)16x(2-x)+(4x-5)^2=0. 2)9y(y+6)-(3y+1)^2=-1.

Answer 1

файл

----------------------------

31da776ac41c8f384a134d233937aef2.doc

Answer 2

$16x(2-x)+(4x-5)^{2}=0$

$32x-16x^{2}+16x^{2}-40x+25=0$

$(16x^{2}-16x^{2})+(32x-40x)+25=0$

$-8x+25=0$

$8x=25$

$x=25:8$

$x=frac{25}{8}=3frac{1}{8}$

===================================================

$9y(y+6)-(3y+1)^{2}=-1$

$9y^{2}+54y-(9y^{2}+6y+1)=-1$

$9y^{2}+54y-9y^{2}-6y-1=-1$

$9y^{2}+54y-9y^{2}-6y-1+1=0$

$(9y^{2}-9y^{2})+(54y-6y)+(-1+1)=0$

$48y=0$

$y=0$

31da776ac41c8f384a134d233937aef2.doc