4x(16-8x+x^2)=7(4-x)помогите с уравнением. Created by dnooooo1. algebra-ru

4x(16-8x+x^2)=7(4-x)помогите с уравнением...

Verified answer

4x (4 - x)^2 - 7(4 - x) = 0
(4 - x)(4x (4 - x) - 7) = 0

1) 4 - x = 0
- x = - 4
x = 4

2) 16x - 4x^2 - 7 = 0
4x^2 - 16x + 7 = 0
D = 144 = 12^2
x2 = ( 16 + 12)/8 = 3,5
x3 = ( 16 - 12)/8 = 0,5

Ответ
0.5; 3.5; 4

2 votes Thanks 6

pentV $4x(16-8x+ x^{2} )=7(4-x); 4x( 4-x)^{2} -7(4-x);(4x(4-x)-7)*$
$*(4-x); \left \{ {{-4 x^{2} +16x-7=0} \atop {4-x=0}} \right. x[tex] x_{1}= \frac{-16+12}{-8} =0.5; x_{2} = \frac{-16-12}{-8} =3.5; x_{3} =4$ =4; D= 16^{2}-4*(-4)*(-7)= 12^{2} [/tex]

0 votes Thanks 4

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