Решите уравнение: 1) (x+3)(x-3)-x(x+4)=0 2)3x(1+12x)-(6x-1)(6x+1)=2,5x...

$(x+3)(x-3)-x(x+4)=0$
(x+3)(x-3) - формула сокращ. умножения; = $x^{2} - 3^{2} = x^{2} - 9$
-x(x+4) = $- x^{2} - 4x$
$x^{2} - 9 - x^{2} - 4x = 0$
$- 9 - 4x = 0$
$- 4x = 9$
$x = - \frac{9}{4} = -2,25$

$3x(1+12x)-(6x-1)(6x+1)=2,5x$
$3x(1+12x) = 3x + 36 x^{2}$
$(6x-1)(6x+1) = 36 x^{2} - 1$
$3+36 x^{2} -36 x^{2} +1 = 2,5x$
$2,5x = 4$
$x = \frac{4}{2 \frac{1}{2} }$
$x = \frac{4}{ \frac{5}{2} }$
$x = 4* \frac{2}{5} = \frac{8}{5} = 1,6$

$(x-5)(x+5)-(2x+1)(x-2)=1-x^2$
$x^{2} -25-(2x^{2} -4x + x - 2) = 1- x^{2}$
$x^{2} -25-2x^{2} +4x - x + 2 = 1- x^{2}$
$-x^{2} -23 +3x = 1- x^{2}$
$-x^{2} + x^{2} +3x = 1 + 23$
$3x = 24$
$x = 8$