\(A=x^2-3x+5=x^2-2.1,5.x+1,5^2+2,75=\left(x-1,5\right)^2+2,75\)
Mà \(\left(x-1,5\right)^2\ge0\Rightarrow\left(x-1,5\right)^2+2,75\ge2,75\)
Dấu "=" xảy ra \(\Leftrightarrow x-1,5=0\Rightarrow x=1,5\)
Vậy GTNN của A là 2,75 khi x = 1,5
\(B=\left(2x+3\right)\left(x-5\right)=2x^2-10x+3x-15=2x^2-7x-15\)
=> \(2B=4x^2-14x-30=\left(2x\right)^2-2.\frac{7}{2}.2x+\frac{49}{4}-42,25=\left(2x-\frac{7}{2}\right)^2-42,25\)
Vì \(\left(2x-\frac{7}{2}\right)^2\ge0\Rightarrow\left(2x-\frac{7}{2}\right)^2-42,25\ge-42,25\Rightarrow2B\ge-42,25\Rightarrow B\ge-21,125\)
Dấu "=" xảy ra \(\Leftrightarrow2x-\frac{7}{2}=0\Rightarrow2x=\frac{7}{2}\Rightarrow x=\frac{7}{4}=1,75\)
Vậy GTNN của B là -21,125 khi x = 1,75
\(A=x^2-3x+5\)
\(=\left(x^2-2.x.\frac{3}{2}-\frac{9}{4}\right)+\frac{29}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{29}{4}\ge\frac{29}{4}\)
Min \(A=\frac{29}{4}\)khi \(x=\frac{3}{2}\)
