\(A=x^2-2x-x+2+3=x\left(x-2\right)-\left(x-2\right)+3=\left(x-2\right).\left(x-1\right)+3\)

Ta có \(x\ge2\Rightarrow x-2\ge0\)

\(x\ge2\Rightarrow x-1\ge1\)

Do đó \(\left(x-2\right).\left(x-1\right)\ge0\)

\(\Rightarrow A=\left(x-2\right)\left(x-1\right)+3\ge3\)

Vậy GTNN của A= 3 khi x-2=0 hay x=2

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

a) \(A=5-8x-x^2=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left[\left(x+4\right)^2-21\right]\)

\(=-\left(x+4\right)^2+21\le21\)

Vậy \(A_{max}=21\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(B=5x-3x^2=-3\left(x^2-\frac{5}{3}x\right)\)

\(=-3\left(x^2-\frac{5}{3}x+\frac{35}{36}-\frac{25}{36}\right)\)

\(=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{25}{36}\right]\)

\(=-3\left[\left(x-\frac{5}{6}\right)^2\right]+\frac{25}{12}\le\frac{25}{12}\)

Vậy \(B_{min}=\frac{25}{12}\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

\(A=2\left|x-5\right|-2015\ge-2015\)

\(Min_A=-2015\Leftrightarrow x=5\)

\(B=205-\left|3x-5\right|\le205\)

\(Max_B=205\Leftrightarrow x=\frac{5}{3}\)

có cách làm củ thể hơn k bạn

đúng rùi đó Nguyễn Văn Tân

rồi đó

\(A=x^2-3x+5=x^2-2.1,5.x+1,5^2+2,75=\left(x-1,5\right)^2+2,75\)

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  \(x=\frac{3}{2}\)