Question

Cho \(a>0\) và \(4a^2+a\sqrt{2}-\sqrt{2}=0\). Chứng minh:

\(\dfrac{a+1}{\sqrt{a^4+a+1}-a^2}=\sqrt{2}\)

Answer 1

\(a^2=\dfrac{\sqrt{2}}{4}\left(1-a\right)\)

\(\Rightarrow a^4=\dfrac{1}{8}\left(1-a\right)^2\)

\(\Rightarrow a^4+a+1=\dfrac{1}{8}\left(1-a\right)^2+a+1=\dfrac{1}{8}\left(a^2+6a+9\right)=\dfrac{1}{8}\left(a+3\right)^2\)

\(\Rightarrow\sqrt{a^4+a+1}-a^2=\sqrt{\dfrac{1}{8}\left(3+a\right)^2}-a^2=\dfrac{\sqrt{2}}{4}\left(a+3\right)-\dfrac{\sqrt{2}}{4}\left(1-a\right)=\dfrac{\sqrt{2}}{2}\left(a+1\right)\)

\(\Rightarrow\dfrac{a+1}{\sqrt{a^4+a+1}-a^2}=\dfrac{a+1}{\dfrac{\sqrt{2}}{2}\left(a+1\right)}=\sqrt{2}\)