$\dfrac{x^{24}+x^{20}+.....+x^4+1}{x^{26} +x^{24}+x^{22}+x^{20}+....x^4+x^2+1} \\
=\dfrac{x^{24}+x^{20}+.....+x^4+1}{x^{24} (x^2+1)+x^{20} (x^2+1)+....+x^4(x^2+1)+(x^2+1)} \\
=\dfrac{x^{24}+x^{20}+.....+x^4+1}{(x^{24} + x^{20}+...+x^4+1)(x^2+1)} \\
=\dfrac{1}{x^2+1}$
1a)
$M= \dfrac{ \left( x+ \dfrac{1}{x} \right) ^6- \left( x^6+ \dfrac{1}{x^6} \right) -2}{ \left( x+ \dfrac{1}{x} \right) ^3+x^3+ \dfrac{1}{x^3}} \; \left( \textrm{ĐKXĐ : }x \ne 0 \right) \\
= \dfrac{ \left( x+ \dfrac{1}{x} \right) ^6- \left( x^6+2+ \dfrac{1}{x^6} \right) }{ \left( x+ \dfrac{1}{x} \right) ^3+x^3+ \dfrac{1}{x^3}} \\
= \dfrac{ \left( x+ \dfrac{1}{x} \right) ^6- \left( x^6+2.x^3. \dfrac1{x^3}+ \dfrac{1}{x^6} \right) }{ \left( x+ \dfrac{1}{x} \right) ^3+x^3+ \dfrac{1}{x^3}} \\
= \dfrac{ \left( x+ \dfrac{1}{x} \right) ^6- \left( x^3+ \dfrac{1}{x^3} \right) ^2}{ \left( x+ \dfrac{1}{x} \right) ^3+x^3+ \dfrac{1}{x^3}} \\
= \dfrac{ \left[ \left( x+ \dfrac{1}{x} \right) ^3- \left( x^3+ \dfrac{1}{x^3} \right) \right] \left[ \left( x+ \dfrac{1}{x} \right) ^3+ \left( x^3+ \dfrac{1}{x^3} \right) \right] }{ \left( x+ \dfrac{1}{x} \right) ^3+ \left( x^3+ \dfrac{1}{x^3} \right) } \\
= \left( x+ \dfrac{1}{x} \right) ^3- \left( x^3+ \dfrac{1}{x^3} \right) \\
= x^3 + 3.x^2. \dfrac1{x} + 3.x. \dfrac1{x^2} + \dfrac1{x^3} - \left( x^3+ \dfrac1{x^3} \right) \\
= 3x+3 \dfrac1{x} \\
=3. \left( \dfrac{x^2+1}{x} \right) $
b) $M=3. \left( x+ \dfrac1{x} \right) $
Áp dụng Cô-si cho 2 số không âm
$M \ge 3.2\sqrt{x. \dfrac1{x}} = 6$
Vậy $\mathrm{Min}_M = 6 \iff x = \dfrac1x = 1$