a) đặt \(A=x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=' xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=-\dfrac{1}{2}\)
b) đặt \(B=2+x-x^2\)
\(=-x^2+x+2\)
\(=-\left(x^2-x-2\right)\)
\(=-\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-2\right]\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)
Vậy \(MAX_B=\dfrac{9}{4}\) khi \(x=\dfrac{1}{2}\)
c) đặt \(C=x^2-4x+1\)
\(=x^2-2\cdot x\cdot2+2^2-4+1\)
\(=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(x=2\)
Vậy \(MIN_c=-3\) khi \(x=2\)
d) đặt \(D=4x^2+4x+11\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2-1+11\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(MIN_D=10\) khi \(x=-\dfrac{1}{2}\)
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