Toán tìm x biết (x-2)^3-(x+2)(x^2-2x+4) +(2x-3) (3x-2) =0 25/11/2021 By Gianna tìm x biết (x-2)^3-(x+2)(x^2-2x+4) +(2x-3) (3x-2) =0
$\text{ Đáp án + lời giải:}$ $(x-2)^3 – (x+2)(x^2-2x+4) + (2x-3)(3x-2) =0$ $\to$ $ x^3 – 6x^2 + 12x – 8 -( x^3 + 8) + 6x^2 – 13x + 6 =0$ $\to$ $ -x -16 + 6=0$ $\to$ $ -x -10= 0$ $\to$ $ x= -10$ Trả lời
`(x-2)^3-(x+2)(x^2-2x+4)+(2x-3)(3x-2)=0` `⇔x^3-6x^2+12x-8-(x^3+8)+6x^2-4x-9x+6=0` `⇔x^3+12x-8-x^3-8-4x-9x+6=0` `⇔-x-10=0` `⇔-x=10` `⇔x=-10` `text{ Vậy x = – 10 }` Trả lời
$\text{ Đáp án + lời giải:}$
$(x-2)^3 – (x+2)(x^2-2x+4) + (2x-3)(3x-2) =0$
$\to$ $ x^3 – 6x^2 + 12x – 8 -( x^3 + 8) + 6x^2 – 13x + 6 =0$
$\to$ $ -x -16 + 6=0$
$\to$ $ -x -10= 0$
$\to$ $ x= -10$
`(x-2)^3-(x+2)(x^2-2x+4)+(2x-3)(3x-2)=0`
`⇔x^3-6x^2+12x-8-(x^3+8)+6x^2-4x-9x+6=0`
`⇔x^3+12x-8-x^3-8-4x-9x+6=0`
`⇔-x-10=0`
`⇔-x=10`
`⇔x=-10`
`text{ Vậy x = – 10 }`