Question

Решите уравнение, плиз!!!!! 1) (x-4)(x^{2} + 4х + 16) + 28 = x^{2} (x-25) 2) (2х+1)(4x^{2} - 2х + 1) - 1 = 1.6x^{2} (5х-2)

Answer 1

$1)(x-4)(x^{2}+4x+16)+28=x^{2}(x-25)\\\\x^{3}+4x^{2}+16x-4x^{2}-16x-64+28=x^{3}-25x^{2}\\\\x^{3}-36-x^{3}+25x^{2}=0\\\\25x^{2}=36\\\\x^{2}=\frac{36}{25}\\\\x_{1}=-\sqrt{\frac{36}{25}}=-\frac{6}{5}=-1,2\\\\x_{2}=\sqrt{\frac{36}{25}}=\frac{6}{5}=1,2\\\\Otvet:\boxed{-1,2;1,2}$

$2)(2x+1)(4x^{2} -2x+1)-1=1,6x^{2}(5x-2)\\\\8x^{3}-4x^{2}+2x+4x^{2} -2x+1-1=8x^{3}-3,2x^{2}\\\\8x^{3}=8x^{3}-3,2x^{2}\\\\3,2x^{2}=0\\\\x=0\\\\Otvet:\boxed{0}$