Để biểu thức \(M=\sqrt{x^4+x^2+1}\) có nghĩa thì \(x^4+x^2+1\ge0\)
\(\Leftrightarrow x^4+2x^2+1-x^2\ge0\)
\(\Leftrightarrow\left(x^2+1\right)^2-x^2\ge0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x+1\right)\ge0\)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
hay \(x^2-x+1\ge\frac{3}{4}\forall x\)
Ta có: \(x^2+x+1\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
hay \(x^2+x+1\ge\frac{3}{4}\forall x\)
Ta có: \(x^2-x+1\ge\frac{3}{4}\forall x\)
\(x^2+x+1\ge\frac{3}{4}\forall x\)
Do đó: \(\left(x^2-x+1\right)\left(x^2+x+1\right)\ge\frac{9}{16}\forall x\)
hay \(x^4+x^2+1>0\forall x\)
Vậy: M xác định được với mọi \(x\in R\)
