a) Ta có: \(A=x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi
\(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTNN của đa thức \(A=x^2-6x+11\) là 2 khi x=3
b) Ta có: \(B=x^2-4x+3\)
\(=x^2-4x+4-1\)
\(=\left(x^2-4x+4\right)-1\)
\(=\left(x-2\right)^2-1\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2-1\ge-1\forall x\)
Dấu '=' xảy ra khi
\(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy: GTNN của đa thức \(B=x^2-4x+3\) là -1 khi x=2
c) Ta có: \(C=x^2+5x\)
\(=x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)
\(=\left(x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\)
Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{-25}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
Vậy: GTNN của đa thức \(C=x^2+5x\) là \(\frac{-25}{4}\) khi \(x=\frac{-5}{2}\)
d) Ta có: \(D=x^2+x+1\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy: GTNN của đa thức \(D=x^2+x+1\) là \(\frac{3}{4}\) khi \(x=\frac{-1}{2}\)
e) Ta có: \(E=4x^2+4x-2\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1-3\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1\right]-3\)
\(=\left(2x+1\right)^2-3\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2-3\ge-3\forall x\)
Dấu '='xảy ra khi
\(\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
Vậy: GTNN của đa thức \(E=4x^2+4x-2\) là -3 khi \(x=\frac{-1}{2}\)
g) Ta có: \(G=x^2-7x\)
\(=x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{14}-\frac{49}{14}\)
\(=\left(x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{4}\right)-\frac{49}{4}\)
\(=\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\)
Ta có: \(\left(x-\frac{7}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\ge\frac{-49}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x-\frac{7}{2}\right)^2=0\Leftrightarrow x-\frac{7}{2}=0\Leftrightarrow x=\frac{7}{2}\)
Vậy: GTNN của đa thức \(G=x^2-7x\) là \(\frac{-49}{4}\) khi \(x=\frac{7}{2}\)
\(A=x^2-6x+11\)
\(A=x^2-2.x.3+3^2-3^2+11\)
\(A=\left(x^2-6x+3^2\right)-3^2+11\)
\(A=\left(x-3\right)^2+2\)
Vì \(\left(x-3\right)^2\ge0\forall x\)
=>\(\left(x-3\right)^2\ge0\ge2\forall x\)
Min A = 2 khi \(\left(x-3\right)^2=0\)
=> \(x-3=0hayx=3\)
Vậy Min A = 2 khi x = 3
\(B=x^2-4x+3\)
\(B=x^2-2.x.2+2^2-2^2+3\)
\(B=\left(x^2-4x+2^2\right)-4+3\)
\(B=\left(x-2\right)^2-1\)
=> \(\left(x-2\right)^2-1\ge0\forall x\)
MIn B = -1 khi \(\left(x-2\right)^2=0\)
=>\(\left(x-2\right)=0hayx=2\)
Vậy Min B = -1 khi x= 2
