Ta có :  

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

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

\(=3\left(x+\frac{1}{x}\right)\ge6\left(x>0\right)\)

\(\Rightarrow Pmin=6\Leftrightarrow x=1\)

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

\(=\frac{\left(x+\frac{1}{x}\right)^6-\left[\left(x^3\right)^2+2x^3\cdot\frac{1}{x^3}+\left(\frac{1}{x^3}\right)^2\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)

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

\(=\frac{\left[\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\right]\left[\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)

\(=\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\ge\left(2\sqrt{x\cdot\frac{1}{x}}\right)^3+2\sqrt{x^3\cdot\frac{1}{x^3}}=8+2=10\)

Dấu "=" khi x = 1

B=\(\frac{3\left(2x^8+5x^6+6x^4+5x^2+2\right)}{x\left(x^2+1\right)\left(2x^4+x^2+2\right)}\)

a/ Đặt: \(x+\frac{1}{x}=a\)

Ta có: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)

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

\(=\left(a^3-3a\right)^2-2\)

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

\(=\frac{a^6-\left(a^3-3a\right)^2+2-2}{a^3+a^3-3a}\)

\(=\frac{\left(a^3+a^3-3a\right)\left(a^3-a^3+3a\right)}{\left(a^3+a^3-3a\right)}=3a\)

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

b/ \(\frac{3x^2+3}{x}=3x+\frac{3}{x}\ge2.3=6\)

Đấu =  xảy ra khi \(x=\frac{1}{x}\Leftrightarrow x=1\)