Question

Giải các phương trình:

\(a,x^4+3x^2-2x+3=0\)

b, \(x^4+x^3-3x^2-4x-4=0\)

Answer 1

b) \(x^4+x^3-3x^2-4x-4=0\)

\(\Leftrightarrow x^4+2x^3-x^3-2x^2-x^2-2x-2x-4=0\)

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

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

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

\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\right]=0\)

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

Vì \(x^2+x+1>0\forall x\)( cách c/m mình nói sau )

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)

Vậy....

Answer 2

Cách chứng minh :

\(x^2+x+1\)

\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

Hay \(x^2+x+1>0\forall x\)( đpcm )