Ta có: M = \(\frac{x^4+x^2+5}{x^4+2x^2+1}\)

M = \(\frac{\left(x^4+2x^2+1\right)-\left(x^2+1\right)+5}{\left(x^2+1\right)^2}\)

M = \(1-\frac{1}{x^2+1}+5\cdot\frac{1}{\left(x^2+1\right)^2}\)

Đặt \(\frac{1}{x^2+1}=y\)

Khi đó, ta có: M = \(1-y+5y^2=5\left(y^2-\frac{1}{5}y+\frac{1}{100}\right)+\frac{19}{20}=5\left(y-\frac{1}{10}\right)^2+\frac{19}{20}\ge\frac{19}{20}\forall y\)

Dấu "=" xảy ra <=> y - 1/10 = 0 <=> y = 1/10 <=> \(\frac{1}{x^2+1}=\frac{1}{10}\) <=> x² + 1 = 10

<=> x² = 9 <=> \(x=\pm3\)

Vậy MinM = 19/20 khi x = 3 hoặc x = -3

Dạng này bạn chỉ cần để ý: \(x^4+2x^2+1=\left(x^2+1\right)^2\) là bình phương của một biểu thức.

Rồi đặt \(x^2+1=y\Rightarrow x^2=y-1\) rồi thay vào M là được!

a/ \(2x^2+8x+1=2\left(x^2+4x+\frac{1}{2}\right)=2\left(x^2+2.2x+4-4+\frac{1}{2}\right)\)

     \(=2\left[\left(x+2\right)^2-\frac{7}{2}\right]=2\left(x+2\right)^2-7\ge-7\)

       Vậy Min A = -7 khi x + 2 = 0 => x = 2

b/ \(2x^2+3x+1=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=2\left(x^2+2.\frac{3}{4}.x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}\right)\)

          \(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

         Vậy Min B = -1/8 khi x + 3/4 = 0 => x = -3/4

\(\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge1\)

Đặt \(\sqrt{x^2-4x+5}=a\Rightarrow a\ge1\)

\(M=2\left(x^2-4x+5\right)+\sqrt{x^2-4x+5}-4\)

\(M=2a^2+a-4=2a^2+3a-2a-3-1\)

\(M=a\left(2a+3\right)-\left(2a+3\right)-1\)

\(M=\left(a-1\right)\left(2a+3\right)-1\)

Do \(a\ge1\Rightarrow\left\{{}\begin{matrix}a-1\ge0\\2a+3>0\end{matrix}\right.\) \(\Rightarrow\left(a-1\right)\left(2a+3\right)\ge0\Rightarrow M\ge-1\)

\(\Rightarrow M_{min}=-1\) khi \(a=1\Leftrightarrow x=2\)

a)Ta thấy:

\(\left(2x+\frac{1}{3}\right)^2\ge0\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^2-\frac{5}{6}\ge0-\frac{5}{6}=-\frac{5}{6}\)

\(\Rightarrow A\ge-\frac{5}{6}\)

Dấu "=" <=>x=-1/6

Vậy MinA=-5/6<=>x=-1/6

b)Ta thấy:\(\hept{\begin{cases}\left|2x+3\right|\\\left|y-\frac{1}{2}\right|\end{cases}\ge}0\)

\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|\ge0\)

\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)

\(\Rightarrow B\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|2x-3\right|=0\\\left|y-\frac{1}{2}\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

Vậy...

\(\text{a)Để C đạt GTNN}\)

\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)

\(\Rightarrow C\ge-10\)

\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)

b)\(\text{Để D đạt GTLN}\)

=>(2x-3)²+5 đạt GTNN

Mà (2x-3)²\(\ge\)5

\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)