a) \(BT=\left(x-4\right)^2+1>0\forall x\)
b, c tương tự
a) Ta có: \(x^2-8x+17\)
\(=x^2-8x+16+1\)
\(=\left(x-4\right)^2+1\)
Ta có: \(\left(x-4\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-4\right)^2+1\ge1>0\forall x\)
hay \(x^2-8x+17>0\forall x\)(đpcm)
b) Ta có: \(4x^2-12x+13\)
\(=\left(2x\right)^2-2\cdot2x\cdot3+9+4\)
\(=\left(2x-3\right)^2+4\)
Ta có: \(\left(2x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x-3\right)^2+4\ge4>0\forall x\)
hay \(4x^2-12x+13>0\forall x\)
c) 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: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
hay \(x^2+x+1>0\forall x\)(đpcm)