\(\Rightarrow3x^4-9x^3+9x^2-27x=0\)
\(\Rightarrow3x^2\left(x^2+3\right)-9x\left(x^2+3\right)=0\)
\(\Rightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(3x^4-9x^3=-9x^2+27x\)
\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)
\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x^3+9x\right)=0\)
\(\Leftrightarrow3x\left(x^2+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-3\left(L\right)\\x=3\end{matrix}\right.\)
Vậy PT có nghiệm là \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
3x4-9x3+9x2-27x=0
(=)3x.(x3-3x2+3x-9)=0
TH1:3x=0
=>x=0
TH2:x3-3x2+3x-9=0
(=)x2.(x-3)+3.(x-3)=0
(=)(x2+3).(x-3)=0
(=)x-3=0
(=)x=3
vậy S={0;3}(=)3x4-9x3=-9x2+27x
3x4 - 9x3 = -9x2 + 27x
⇔ 3x4 - 9x3 + 9x2 - 27x = 0
⇔ 3x( x3 - 3x2 + 3x - 9 ) = 0
⇔ 3x [( x3 - 3x2 ) + ( 3x - 9 )] = 0
⇔ 3x[ x2( x - 3 ) + 3( x - 3 ) ] = 0
⇔ 3x( x - 3 )( x2 + 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-3=0\\x^2+3=0\left(k.TM\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy x = 0 hoặc x = 3
* k.TM là không thỏa mãn nhé
