Câu 1:
\(A=x^2+10x+27=\left(x^2+10x+25\right)+2\)
\(=\left(x+5\right)^2+2\ge2\forall x\)
vậy : Min A = 2 khi x + 5 =0 => x =-5
\(B=x^2+x+7=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{27}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\forall x\)
Vậy Min B = \(\dfrac{27}{4}\) khi \(x+\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(C=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Vậy Min C = \(\dfrac{11}{4}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
Câu 2:
\(A=-x^2+2x+2=-\left(x^2-2x+1\right)+3\)
\(=-\left(x-1\right)^2+3\le3\forall x\)
Vậy Max A = 3 khi x-1=0=> x =1
\(B=-x^2+8x+17=-\left(x^2-8x+16\right)+33\)\(=-\left(x-4\right)^2+33\le33\forall x\)
Vậy Max B = 33 khi x - 4 =0 => x = 4
