Câu hỏi của Tú Nguyễn

Question

Tìm x biết (x+3)^3-3.(3x+1)^2+(2x+1).(4x^2-2x+1)=54

Đức Phạm · Accepted Answer

$\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54$

$\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54$

$\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54$

$\Leftrightarrow9x^3-18x^2+9x-29=0$

$\Leftrightarrow x=2,208024627$

Câu trả lời:

$\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54$

$\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54$

$\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54$

$\Leftrightarrow9x^3-18x^2+9x-29=0$

$\Leftrightarrow x=2,208024627$