Câu hỏi của erza sarlet

Question

Tìm x biết:

(x-3)(x^2+3x+9)-(3x-17)=x^3-12

T.Thùy Ninh · Accepted Answer

$\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12$

$\Leftrightarrow x^3-27-3x+17-x^3+12=0$

$\Leftrightarrow2-3x=0$

$\Leftrightarrow3x=2\Rightarrow x=\dfrac{2}{3}$Câu trả lời: $\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12$

$\Leftrightarrow x^3-27-3x+17-x^3+12=0$

$\Leftrightarrow2-3x=0$

$\Leftrightarrow3x=2\Rightarrow x=\dfrac{2}{3}$

DƯƠNG PHAN KHÁNH DƯƠNG · Answer

$\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12$

$\Leftrightarrow x^3-27-3x+17-x^3+12=0$

$\Leftrightarrow-3x+2=0$

$\Rightarrow x=\dfrac{2}{3}$

Vậy .............Câu trả lời: $\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12$

$\Leftrightarrow x^3-27-3x+17-x^3+12=0$

$\Leftrightarrow-3x+2=0$

$\Rightarrow x=\dfrac{2}{3}$

Vậy .............

Nguyễn Thị Hồng Nhung · Answer

$\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12$

$x^3-3^3-3x+17=x^3-12$

$-3x=-12+27-17=-2$

=>$x=\dfrac{2}{3}$

Vậy...Câu trả lời: $\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12$

$x^3-3^3-3x+17=x^3-12$

$-3x=-12+27-17=-2$

=>$x=\dfrac{2}{3}$

Vậy...

Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)