\(Q=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

Q=(x+1)² 
GTNN là Q=0
xảy ra khi (x+1)²=0
                  x+1=0
                  x     =-1

Biểu thức này chỉ có GTLN thôi.

\(A=\frac{3}{2x^2+x+1}=\frac{3}{2\left(x^2+\frac{1}{2}x+\frac{1}{2}\right)}=\frac{3}{2\left[\left(x+\frac{1}{4}\right)^2+\frac{7}{16}\right]}=\frac{3}{2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}}\le\frac{3}{\frac{7}{8}}=\frac{24}{7}\)

GTLN của A là \(\frac{24}{7}\) khi \(x+\frac{1}{4}=0\Rightarrow x=-\frac{1}{4}\)

x²+x+1=x²+2x+1-x-1+1=(x+1)²-(x+1)+1

x²+2x+1=(x+1)²

đặt x+1=y , A=(y²-y+1)/y²=1-1/y+1/y²

tiếp tục đặt 1/y=z ta đc A=z²-z+1=(z-1/2)²+3/4 >= 3/4

minA=3/4 dấu "=" khi z=1/2<=>y=2<=>x=1

Bài 1: \(\sqrt{x^2+2x+5}=\sqrt{\left(x^2+2x+1\right)+4}\)

\(=\sqrt{\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Dấu "=" xảy ra khi \(x=-1\)

Vậy...

Bài 2:

\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=\left|2x-1\right|+\left|2x-3\right|\)\(=\left|2x-1\right|+\left|3-2x\right|\)

\(\ge\left|2x-1+3-2x\right|=2\)

Dấu "=" xảy ra khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

Vạy....

\(A=\dfrac{1}{2}\left(x-3\right)^2+10\ge10\\ A_{min}=10\Leftrightarrow x-3=0\Leftrightarrow x=3\)

\(A=\dfrac{1}{2}\left(x-3\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi x=3