Question

[Vũ Phạm Khánh Ngọc]

Tìm x biết x.(x+2).(x+4).(x+6)= 9

Answer 1

[Kiyotaka Ayanokoji]

Trả lời 

\(x.\left(x+2\right).\left(x+4\right).\left(x+6\right)=9\)

\(\Leftrightarrow\left[x.\left(x+6\right)\right].\left[\left(x+2\right).\left(x+4\right)\right]=9\)

\(\Leftrightarrow\left(x^2+6x\right).\left(x^2+6x+8\right)=9\)

Đặt \(x^2+6x=t\) ta có 

\(t.\left(t+8\right)=9\)

\(\Leftrightarrow t^2+8t-9=0\)

\(\Leftrightarrow t^2-t+9t-9=0\)

\(\Leftrightarrow t.\left(t-1\right)+9.\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right).\left(t+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t+9=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=1\\t=-9\end{cases}}\)

TH1 \(t=1\)

\(\Rightarrow x^2+6x=1\)

\(\Leftrightarrow x^2+6x-1=0\)

\(\Leftrightarrow x^2+6x+9-10=0\)

\(\Leftrightarrow\left(x+3\right)^2=10=\left(\pm\sqrt{10}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3+\sqrt{10}\\x=-3-\sqrt{10}\end{cases}}\)

TH2: \(t=-9\)

\(\Rightarrow x^2+6x=-9\)

\(\Leftrightarrow x^2+6x+9=0\)

\(\Leftrightarrow\left(x+3\right)^2=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy \(x\in\left\{-3+\sqrt{10};-3-\sqrt{10};-3\right\}\)

Answer 2

[๖²⁴ʱんuﾘ ｲú❄✎﹏]

\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)

\(\Leftrightarrow x^4+12x^3+44x^2+48x=9\)

\(\Leftrightarrow x^4+12x^3+44x^2+48x-9=0\)

\(\Leftrightarrow\left(x^3+9x^2+17x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)

TH1 : Ta có : \(6^2-4.\left(-1\right)=36+4=40>0\)Suy ra : \(x_1=\frac{-6-\sqrt{40}}{2};x_2=\frac{-6+\sqrt{40}}{2}\)

TH2 : \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)