Question

Привидите уравнение к виду ax^2+bx+c=0
a)x^2-x=12
b)(x+5)^2=(3-2x)^2
B)4х(1-х)=х^2-5х-2
Г) (2х-1)^2+4х=1

Answer 1

а)
${x }^{2} - x - 12 = 0$
б)
${(x + 5)}^{2} = {(3 - 2x)}^{2} \\ {x}^{2} + 10x + 25 = 9 - 12x + {4x}^{2} \\ {x}^{2} - {4x}^{2} + 10x + 12x + 25 - 9 = 0 \\ - 3 {x}^{2} + 24x + 16 = 0 \\ {3x}^{2} - 24x - 16 = 0$
в)
$4x(1 - x) = {x}^{2} - 5x - 2 \\ 4x - {4x}^{2} = {x}^{2} - 5x - 2 \\ - 4 {x}^{2} - {x}^{2} + 4x + 5x + 2 = 0 \\ - 5 {x}^{2} + 9x + 2 = 0 \\ 5 {x }^{2} - 9x - 2 = 0$

г)
$(2x - 1) ^{2} + 4x = 1 \\ {4x}^{2} - 4x + 1 + 4x - 1 = 0 \\ 4 {x }^{2} = 0$