Question

Tìm x: ( x-2 )³ - ( x+2 ) ( x² -2x+4 ) + ( 2x -3 ) ( 3x -2 )=0

Answer 1

mk làm lun nha

[(x-2)^3-(x+2)^3]+6x^2-4x-9x+6=0

(x-2-x-2)[(x-2)^2+(x-2)(x+2)+(x+2)^2]+6x^2-4x-9x+6=0

-4[x^2-4x+4+x^2-4+x^2+4x+4]+6x^2-4x-9x+6=0

-4[3x^2+4]+6x^2-4x-9x+6=0

-12x^2-16+6x^2-4x-9x+6=0

-6x^2-13x-10=0

x=-13/12

**** cho mk nha

Answer 2

mình sai rồi, xin sửa lại:

x^3 - 6x^2 + 12x - 8 - x^3 - 8 + 6x^2 - 4x - 9x + 6 = 0

-x - 10 = 0

-x = 10

\(\Rightarrow\) x = -10