\(\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)+x^3+\dfrac{1}{x^3}}\)
\(=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+2+\dfrac{1}{x^6}\right)}{\left(x+\dfrac{1}{x}\right)+\left(x^3+\dfrac{1}{x^3}\right)}\)
\(=\dfrac{\left[\left(x+\dfrac{1}{x}\right)^3\right]^2-\left(x^3+\dfrac{1}{x^3}\right)^2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\)
\(=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)\)
\(=3x+\dfrac{3}{x}\)
\(=3\left(x+\dfrac{1}{x}\right)\)
Áp dụng bất đẳng thức \(x+\dfrac{1}{x}\ge2\forall x>0\)
\(\Rightarrow3\left(x+\dfrac{1}{x}\right)\ge6\)
\(\Rightarrowđpcm\)
\(S=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\)
☘ Đặt \(M=\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)\)
\(=\left(x+\dfrac{1}{x}\right)^3+\left[\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\right]\)
\(=2\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\)
☘ Đặt \(N=\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2\)
\(=\left(x+\dfrac{1}{x}\right)^6-\left[\left(x^2+\dfrac{1}{x^2}\right)^3-3\left(x^2+\dfrac{1}{x^2}\right)\right]-2\)
\(=\left(x+\dfrac{1}{x}\right)^6-\left\{\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^3-3\left[\left(x+\dfrac{1}{x}\right)^2-2\right]\right\}-2\)
\(=\left(x+\dfrac{1}{x}\right)^6-\left[\left(x+\dfrac{1}{x}\right)^6-6\left(x+\dfrac{1}{x}\right)^4+12\left(x+\dfrac{1}{x}\right)^2-8-3\left(x+\dfrac{1}{x}\right)^2+6\right]-2\)
\(=6\left(x+\dfrac{1}{x}\right)^4-9\left(x+\dfrac{1}{x}\right)^2\)
☘ Đặt \(x+\dfrac{1}{x}=a\)
\(\Rightarrow S=\dfrac{6a^4-9a^2}{2a^3-3a}=3a\)
Áp dụng bất đẳng thức AM - GM
\(\Rightarrow a=x+\dfrac{1}{x}\ge2\)
\(\Rightarrow S\ge6\) (đpcm)
Vậy \(S\ge6\Leftrightarrow x=1\)
♬♫♪ Cách này hơi dài. Nhưng thật sự, chưa nghĩ được cách khác ngắn hơn. Thông cảm nhé.
