Question

$(3x + {2})^{2} = 12x + 40$
$(x - 1) ( x+ 1)=2(x - 2) (x+2)$
$48-3( {x}^{2} - 5)²=0$
РЕШИТЕ ПЛИЗ ДАЮ 80 БАЛЛОВ​

Answer 1

Ответ:

1) $(3x + 2) {}^{2} = 12x + 40 \\ {9x}^{2} + 12x + 4 = 12x + 40 \\ {9x}^{2} + 4 = 40 \\ 9x {}^{2} = 40 - 4 \\ {9x}^{2} = 36$

${x}^{2} = 4 \\ x = + 2 \\ x = - 2 \\ x = 2$

2) $(x - 1)(x + 1) = 2(x - 2)(x + 2) \\ {x}^{2} - 1 = 2( {x}^{2} - 4) \\ {x}^{2} - 1 = 2x {}^{2} - 8 \\ {x}^{2} - 2x {}^{2} = - 8 + 1$

${ - x}^{2} = - 7 \\ {x}^{2} = 7 \\ x = + \sqrt{7} \\ x = - \sqrt{7} \\ x = \sqrt{7}$

3)

$48 - 3( {x}^{2} - 5) {}^{2} = 0 \\ - 3(x {}^{2} - 5) {}^{2} = - 48$

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

${x}^{2} - 5 = + 4$

${x}^{2} - 5 = - 4$

${x}^{2} - 5 = 4$

$x = - 1 \\ x = 1 \\ x = - 3 \\ x = 3$