Question

решите уравнение:
5t^2-4t=0
(5x-2)(2x+1)=(5x-2)(4x-1)
(2x+31)+(4-5x)=6x-1
15x^2-(5x+2)(3x-4)=1

Answer 1

5t^2-4t=0
t(5t-4)=0
t=0
5t-4=0
5t=4
t=0,8

(5x-2)(2x+1)=(5x -2)(4x-1)|:(5x-2)
5x -2 не = 0
5x не = 2
х не = 0,4
2х+1=4х-1
2х-4х=-1-1
-2х=-2
х =-1

2х+31+4-5х=6х-1
2х-5х-6х=-1-31-4
-9х=-36
х=4

15х^2-(15х^2-20х+6х-8)=1
15х^2-15х^2+14х+8=1
14х=-7
х=-1/2=-0,5