\(x^3-6x^2-x+6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\\x=-1\end{matrix}\right.\)

Vậy ................................

Bài nào bn

\(x^3-6x^2+11x-6=0\\ \Leftrightarrow\left(x^3-x^2\right)-\left(5x^2-5x\right)+\left(6x-6\right)=0\\ \Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

\(x^3+4x^2+x+6=0\)

\(\Leftrightarrow\text{ (x + 3).(x + 2).(x - 1) = 0 }\)

<=>

Tự làm nhé mk nhẩm ko nhầm là dậy :D

\(x^3+4x^2+x-6=x^3-x^2+5x^2-5x+6x-6=x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x^2+5x+6\right)=\left(x-1\right)\left(x^2+2x+3x+6\right)=\left(x-1\right)\left(x+2\right)\left(x+3\right)\)

chúc bạn học tốt

x³-6x²+11x-6=0

⇔x³-x²-5x²+5x+6x-6=0

⇔(x³-x²)-(5x²-5x)+(6x-6)=0

⇔x²(x-1)-5x(x-1)+6(x-1)=0

⇔(x-1)(x²-5x+6)=0

⇔(x-1)(x²-2x-3x+6)=0

⇔(x-1)[(x²-2x)-(3x-6)]=0

⇔(x-1)[x(x-2)-3(x-2)]=0

⇔(x-1)(x-2)(x-3)=0

=>\(\left\{{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy S={1;2;3}

x³-4x²+x+6=0

⇔x³+x²-5x²-5x+6x+6=0

⇔(x³+x²)-(5x²-5x)+(6x+6)=0

⇔x²(x+1)-5x(x+1)+6(x+1)=0

⇔(x+1)(x²-5x+6)=0

⇔(x+1)(x²-2x-3x+6)=0

⇔(x+1)[(x²-2x)-(3x-6)]=0

⇔(x+1)[x(x-2)-3(x-2)]=0

⇔(x+1)(x-2)(x-3)=0

⇔\(\left[{}\begin{matrix}x+1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=3\end{matrix}\right.\)

vậy S={-1;2;3}

Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\)

\(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

Đặt \(x+\frac{1}{x}=a\) (\(\left|a\right|\ge2\)) \(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(6\left(a^2-2\right)+7a-36=0\)

\(\Leftrightarrow6a^2+7a-48=0\)

Nghiệm xấu