Question

Решите уравнение (1-2х)(4х^2+2x+1)=8(1-x^2)(x+2)

Answer 1

(1-2х)(4х²+2х+1)=8(1-х²)(х+2)
4х²+2х+1-8х³-4х²-2х=8х+16-8х³-16х²
-8х³+8х³+16х²²-8х-16+1=0
16х²-8х-15=0
D=8²-4·16·(-15)=1024      √D=32
х1=5/4
х2=-3/4

Answer 2

$(1-2x)(4 x^{2} +2x+1)=8(1- x^{2} )(x+2)\\1-8 x^{3} =8(x+2- x^{3} -2 x^{2} )\\1-8 x^{3} =8x+16-8 x^{3} -16 x^{2} \\1-8 x^{3} -8x-16+8 x^{3} +16 x^{2} =0\\-15-8x+16 x^{2} =0\\16 x^{2} -8x-15=0\\D=64+950=1024\\ \sqrt{D} =32\\ x_{1} = \frac{8+32}{32} = \frac{40}{32} =1.25\\ x_{2} = \frac{8-32}{32} = \frac{-24}{32} =-0.75$