a) \(3x^2-5x+7=3\left(x^2-\frac{5}{3}x+\frac{7}{3}\right)\) \(=3\left(x^2-2\cdot x\cdot\frac{5}{6}+\frac{25}{36}+\frac{59}{36}\right)=3\left(x-\frac{5}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}\)
Dấu bằng xảy ra \(\Leftrightarrow x-\frac{5}{6}=0\) \(\Leftrightarrow x=\frac{5}{6}\)
Vậy GTNN của biểu thức là \(\frac{59}{12}\) khi \(x=\frac{5}{6}\)
b) \(\left(x-1\right)\left(x-3\right)+11=x^2-4x+14\) \(=x^2-2\cdot2\cdot x+4+10=\left(x-2\right)^2+10\ge10\)
Dấu bằng xảy ra \(\Leftrightarrow x=2\)
Vậy GTNN là \(10\) khi \(x=2\)
c) \(\left(x-3\right)^2+\left(x-2\right)^2=2x^2-10x+13=2\left(x^2-5x+\frac{13}{2}\right)\) \(=2\left(x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{1}{4}\right)=2\left(x-\frac{5}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow x=\frac{5}{2}\)
Vậy GTNN là \(\frac{1}{2}\) khi \(x=\frac{5}{2}\)
