Question

[Khánh Ko Ổn]

1. x^4+x^2-2=0;         2. x^3+3x^2+6x+4=0;         3. x^3-6x^2+8x=0;         4. x^4-8x^3-9x^2=0                                  Giúp với (;~;)

Answer 1

[Minh Hồng]

1/ \(x^4+x^2-2=0\)

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

2/ \(x^3+3x^2+6x+4=0\)

\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)

\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))

\(\Leftrightarrow x=-1\).

3/ \(x^3-6x^2+8x=0\)

\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)

4/ \(x^4-8x^3-9x^2=0\)

\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)