Question

Giải bất phương trình: |x³ + 1| ≥ x + 1.

Answer 1

Vì \(VT\ge0\Rightarrow x+1\ge0\)

\(\Rightarrow x\ge-1\)

\(BPT\Leftrightarrow\left\{{}\begin{matrix}x^3+1=x+1\\x^3+1=-\left(x+1\right)\end{matrix}\right.\)

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

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

Vì \(x^2-x+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\) nên

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

Answer 2

Vì 

BPT⇔{x3+1=x+1x3+1=−(x+1)BPT⇔{x3+1=x+1x3+1=−(x+1)

⇔{x3−x=0x3+1+x+1=0⇔{x3−x=0x3+1+x+1=0

⇔{x(x−1)(x+1)=0(x+1)(x2−x+2)=0⇔{x(x−1)(x+1)=0(x+1)(x2−x+2)=0(2)

Vì  nên

(2)⇔⎡⎢⎣x=0(TM)x=1(TM)x=−1(TM)